46. We add some theorems concerning rational functions of complex numbers. A rational function of the complex variable $$z$$ is defined exactly as is a rational function of a real variable $$x$$, viz. as the quotient of two polynomials in $$z$$.

Any rational function $$R(z)$$ can be reduced to the form $$X + Yi$$, where $$X$$ and $$Y$$ are rational functions of $$x$$ and $$y$$ with real coefficients.

In the first place it is evident that any polynomial $$P(x + yi)$$ can be reduced, in virtue of the definitions of addition and multiplication, to the form $$A + Bi$$, where $$A$$ and $$B$$ are polynomials in $$x$$ and $$y$$ with real coefficients. Similarly $$Q(x + yi)$$ can be reduced to the form $$C + Di$$. Hence $R(x + yi) = P(x + yi)/Q(x + yi)$ can be expressed in the form \begin{aligned} (A + Bi)/(C + Di) &= (A + Bi) (C – Di)/(C + Di) (C – Di)\\ &= \frac{AC + BD}{C^{2} + D^{2}} + \frac{BC – AD}{C^{2} + D^{2}} i,\end{aligned} which proves the theorem.

If $$R(x + yi) = X + Yi$$, $$R$$ denoting a rational function as before, but with real coefficients, then $$R(x – yi) = X – Yi$$.

In the first place this is easily verified for a power $$(x + yi)^{n}$$ by actual expansion. It follows by addition that the theorem is true for any polynomial with real coefficients. Hence, in the notation used above, $R(x – yi) = \frac{A – Bi}{C – Di} = \frac{AC + BD}{C^{2} + D^{2}} – \frac{BC – AD}{C^{2} + D^{2}}i,$ the reduction being the same as before except that the sign of $$i$$ is changed throughout. It is evident that results similar to those of Theorems 1 and 2 hold for functions of any number of complex variables.

The roots of an equation $a_{0}z^{n} + a_{1}z^{n-1} + \dots + a_{n} = 0,$ whose coefficients are real, may, in so far as they are not themselves real, be arranged in conjugate pairs.

For it follows from Theorem 2 that if $$x + yi$$ is a root then so is $$x – yi$$. A particular case of this theorem is the result (§ 43) that the roots of a quadratic equation with real coefficients are either real or conjugate.

This theorem is sometimes stated as follows: in an equation with real coefficients complex roots occur in conjugate pairs. It should be compared with the result of 7, which may be stated as follows: in an equation with rational coefficients irrational roots occur in conjugate pairs.1

Examples XXI

1. Prove theorem (6) of § 45 directly from the definitions and without the aid of geometrical considerations.

[First, to prove that $$|z + z’| \leq |z| + |z’|$$ is to prove that $(x + x’)^{2} + (y + y’)^{2} \leq \{\sqrt{x^{2} + y^{2}} + \sqrt{x’^{2} + y’^{2}}\}^{2}.$ The theorem is then easily extended to the general case.]

2. The one and only case in which $|z| + |z’| + \dots = |z + z’ + \dots|,$ is that in which the numbers $$z$$, $$z’$$, … have all the same amplitude. Prove this both geometrically and analytically.

3. The modulus of the sum of any number of complex numbers is not less than the sum of their real (or imaginary) parts.

4. If the sum and product of two complex numbers are both real, then the two numbers must either be real or conjugate.

5. If $a + b\sqrt{2} + (c + d \sqrt{2})i = A + B\sqrt{2} + (C + D\sqrt{2})i,$ where $$a$$, $$b$$, $$c$$, $$d$$, $$A$$, $$B$$, $$C$$, $$D$$ are real rational numbers, then $a = A,\quad b = B,\quad c = C,\quad d = D.$

6. Express the following numbers in the form $$A + Bi$$, where $$A$$ and $$B$$ are real numbers: $(1 + i)^{2},\quad \left(\frac{1 + i}{1 – i}\right)^{2},\quad \left(\frac{1 – i}{1 + i}\right)^{2},\quad \frac{\lambda + \mu i}{\lambda – \mu i},\quad \left(\frac{\lambda + \mu i}{\lambda – \mu i}\right)^{2} – \left(\frac{\lambda – \mu i}{\lambda + \mu i}\right)^{2},$ where $$\lambda$$ and $$\mu$$ are real numbers.

7. Express the following functions of $$z = x + yi$$ in the form $$X + Yi$$, where $$X$$ and $$Y$$ are real functions of $$x$$ and $$y$$: $$z^{2}$$, $$z^{3}$$, $$z^{n}$$, $$1/z$$, $$z + (1/z)$$, $$(\alpha + \beta z)/(\gamma + \delta z)$$, where $$\alpha$$, $$\beta$$, $$\gamma$$, $$\delta$$ are real numbers.

8. Find the moduli of the numbers and functions in the two preceding examples.

9. The two lines joining the points $$z = a$$, $$z = b$$ and $$z = c$$, $$z = d$$ will be perpendicular if $\operatorname{am}\left(\frac{a – b}{c – d}\right) = \pm\tfrac{1}{2} \pi,$ i.e. if $$(a – b)/(c – d)$$ is purely imaginary. What is the condition that the lines should be parallel?

10. The three angular points of a triangle are given by $$z = \alpha$$, $$z = \beta$$, $$z = \gamma$$, where $$\alpha$$, $$\beta$$, $$\gamma$$ are complex numbers. Establish the following propositions:

(i) the centre of gravity is given by $$z = \frac{1}{3}(\alpha + \beta + \gamma)$$;

(ii) the circum-centre is given by $$|z – \alpha| = |z – \beta| = |z – \gamma|$$;

(iii) the three perpendiculars from the angular points on the opposite sides meet in a point given by $\Re\left(\frac{z – \alpha}{\beta – \gamma}\right) = \Re\left(\frac{z – \beta}{\gamma – \alpha}\right) = \Re\left(\frac{z – \gamma}{\alpha – \beta}\right) = 0;$

(iv) there is a point $$P$$ inside the triangle such that $CBP = ACP = BAP = \omega,$ and $\cot\omega = \cot A + \cot B + \cot C.$

[To prove (iii) we observe that if $$A$$, $$B$$, $$C$$ are the vertices, and $$P$$ any point $$z$$, then the condition that $$AP$$ should be perpendicular to $$BC$$ is (Ex. 9) that $$(z – \alpha)/(\beta – \gamma)$$ should be purely imaginary, or that $\Re(z – \alpha) \Re(\beta – \gamma) + \Im(z – \alpha) \Im(\beta – \gamma) = 0.$ This equation, and the two similar equations obtained by permuting $$\alpha$$, $$\beta$$, $$\gamma$$ cyclically, are satisfied by the same value of $$z$$, as appears from the fact that the sum of the three left-hand sides is zero.

To prove (iv), take $$BC$$ parallel to the positive direction of the axis of $$x$$. Then2 $\gamma – \beta = a,\quad \alpha – \gamma = – b\operatorname{Cis}(-C),\quad \beta – \alpha = – c\operatorname{Cis} B.$

We have to determine $$z$$ and $$\omega$$ from the equations $\frac{(z – \alpha)(\beta_{0} – \alpha_{0})} {(z_{0} – \alpha_{0})(\beta – \alpha)} = \frac{(z – \beta)(\gamma_{0} – \beta_{0})} {(z_{0} – \beta_{0})(\gamma – \beta)} = \frac{(z – \gamma)(\alpha_{0} – \gamma_{0})} {(z_{0} – \gamma_{0})(\alpha – \gamma)} = \operatorname{Cis} 2\omega,$ where $$z_{0}$$, $$\alpha_{0}$$, $$\beta_{0}$$, $$\gamma_{0}$$ denote the conjugates of $$z$$, $$\alpha$$, $$\beta$$, $$\gamma$$.

Adding the numerators and denominators of the three equal fractions, and using the equation $i\cot\omega = (1 + \operatorname{Cis} 2\omega)/(1 – \operatorname{Cis} 2\omega),$ we find that $i\cot\omega = \frac{(\beta – \gamma)(\beta_{0} – \gamma_{0}) + (\gamma – \alpha)(\gamma_{0} – \alpha_{0}) + (\alpha – \beta)(\alpha_{0} – \beta_{0})} {\beta\gamma_{0} – \beta_{0}\gamma + \gamma\alpha_{0} – \gamma_{0}\alpha + \alpha\beta_{0} – \alpha_{0}\beta}.$ From this it is easily deduced that the value of $$\cot\omega$$ is $$(a^{2} + b^{2} +c^{2})/4\Delta$$, where $$\Delta$$ is the area of the triangle; and this is equivalent to the result given.

To determine $$z$$, we multiply the numerators and denominators of the equal fractions by $$(\gamma_{0} – \beta_{0})/(\beta – \alpha)$$, $$(\alpha_{0} – \gamma_{0})/(\gamma – \beta)$$, $$(\beta_{0} – \alpha_{0})/(\alpha – \gamma)$$, and add to form a new fraction. It will be found that $z = \frac{a\alpha \operatorname{Cis} A + b\beta \operatorname{Cis} B + c\gamma \operatorname{Cis} C} {a\operatorname{Cis} A + b\operatorname{Cis} B + c\operatorname{Cis} C}.]$

11. The two triangles whose vertices are the points $$a$$, $$b$$, $$c$$ and $$x$$, $$y$$, $$z$$ respectively will be similar if $\begin{vmatrix} 1 & 1 & 1\\ a & b & c \\ x & y & z \end{vmatrix} = 0$

12. Deduce from the last example that if the points $$x$$, $$y$$, $$z$$ are collinear then we can find real numbers $$\alpha$$, $$\beta$$, $$\gamma$$ such that $$\alpha + \beta + \gamma = 0$$ and $$\alpha x + \beta y + \gamma z = 0$$, and conversely (cf. Exs. xx. 4). [Use the fact that in this case the triangle formed by $$x$$, $$y$$, $$z$$ is similar to a certain line-triangle on the axis $$OX$$, and apply the result of the last example.]

13. The general linear equation with complex coefficients. The equation $$\alpha z + \beta = 0$$ has the one solution $$z = -(\beta/\alpha)$$, unless $$\alpha = 0$$. If we put $\alpha = a + Ai,\quad \beta = b + Bi,\quad z = x + yi,$ and equate real and imaginary parts, we obtain two equations to determine the two real numbers $$x$$ and $$y$$. The equation will have a real root if $$y = 0$$, which gives $$ax + b = 0$$, $$Ax + B = 0$$, and the condition that these equations should be consistent is $$aB – bA = 0$$.

14. The general quadratic equation with complex coefficients. This equation is $(a + Ai)z^{2} + 2(b + Bi)z + (c + Ci) = 0.$

Unless $$a$$ and $$A$$ are both zero we can divide through by $$a + iA$$. Hence we may consider $\begin{equation*} z^{2} + 2(b + Bi)z + (c + Ci) = 0 \tag{1} \end{equation*}$ as the standard form of our equation. Putting $$z = x + yi$$ and equating real and imaginary parts, we obtain a pair of simultaneous equations for $$x$$ and $$y$$, viz. $x^{2} – y^{2} + 2(bx – By) + c = 0,\quad 2xy + 2(by + Bx) + C = 0.$

If we put $x + b = \xi,\quad y + B = \eta,\quad b^{2} – B^{2} – c = h,\quad 2bB – C = k,$ these equations become $\xi^{2} – \eta^{2} = h,\quad 2\xi\eta = k.$

Squaring and adding we obtain $\xi^{2} + \eta^{2} = \sqrt{h^{2} + k^{2}},\quad \xi = \pm\sqrt{\tfrac{1}{2}\{\sqrt{h^{2} + k^{2}} + h\}},\quad \eta = \pm\sqrt{\tfrac{1}{2}\{\sqrt{h^{2} + k^{2}} – h\}}.$ We must choose the signs so that $$\xi\eta$$ has the sign of $$k$$: i.e. if $$k$$ is positive we must take like signs, if $$k$$ is negative unlike signs.

Conditions for equal roots. The two roots can only be equal if both the square roots above vanish, i.e. if $$h = 0$$, $$k = 0$$, or if $$c = b^{2} – B^{2}$$, $$C = 2bB$$. These conditions are equivalent to the single condition $$c + Ci = (b + Bi)^{2}$$, which obviously expresses the fact that the left-hand side of (1) is a perfect square.

Condition for a real root. If $$x^{2} + 2(b + Bi) x + (c + Ci) = 0$$, where $$x$$ is real, then $$x^{2} + 2bx + c = 0$$, $$2Bx + C = 0$$. Eliminating $$x$$ we find that the required condition is $C^{2} – 4bBC + 4cB^{2} = 0.$

Condition for a purely imaginary root. This is easily found to be $C^{2} – 4bBC – 4b^{2}c = 0.$

Conditions for a pair of conjugate complex roots. Since the sum and the product of two conjugate complex numbers are both real, $$b + Bi$$ and $$c + Ci$$ must both be real, i.e. $$B = 0$$, $$C = 0$$. Thus the equation (1) can have a pair of conjugate complex roots only if its coefficients are real. The reader should verify this conclusion by means of the explicit expressions of the roots. Moreover, if $$b^{2}\geq c$$, the roots will be real even in this case. Hence for a pair of conjugate roots we must have $$B = 0$$, $$C = 0$$, $$b^{2} < c$$.

15. The Cubic equation. Consider the cubic equation $z^{3} + 3Hz + G = 0,$ where $$G$$ and $$H$$ are complex numbers, it being given that the equation has (a) a real root, (b) a purely imaginary root, (c) a pair of conjugate roots. If $$H = \lambda + \mu i$$, $$G = \rho + \sigma i$$, we arrive at the following conclusions.

(a) Conditions for a real root. If $$\mu$$ is not zero, then the real root is $$-\sigma/3\mu$$, and $$\sigma^{3} + 27\lambda\mu^{2}\sigma – 27\mu^{3}\rho = 0$$. On the other hand, if $$\mu = 0$$ then we must also have $$\sigma = 0$$, so that the coefficients of the equation are real. In this case there may be three real roots.

(b) Conditions for a purely imaginary root. If $$\mu$$ is not zero then the purely imaginary root is $$(\rho/3\mu)i$$, and $$\rho^{3} – 27\lambda\mu^{2}\rho – 27\mu^{3}\sigma = 0$$. If $$\mu = 0$$ then also $$\rho = 0$$, and the root is $$yi$$, where $$y$$ is given by the equation $$y^{3} – 3\lambda y – \sigma = 0$$, which has real coefficients. In this case there may be three purely imaginary roots.

(c) Conditions for a pair of conjugate complex roots. Let these be $$x + yi$$ and $$x – yi$$. Then since the sum of the three roots is zero the third root must be $$-2x$$. From the relations between the coefficients and the roots of an equation we deduce $y^{2} – 3x^{2} = 3H,\quad 2x(x^{2} + y^{2}) = G.$ Hence $$G$$ and $$H$$ must both be real.

In each case we can either find a root (in which case the equation can be reduced to a quadratic by dividing by a known factor) or we can reduce the solution of the equation to the solution of a cubic equation with real coefficients.

16. The cubic equation $$x^{3} + a_{1}x^{2} + a_{2}x + a_{3} = 0$$, where $$a_{1} = A_{1} + A_{1}’i$$, …, has a pair of conjugate complex roots. Prove that the remaining root is $$-A_{1}’a_{3}/A_{3}’$$, unless $$A_{3}’ = 0$$. Examine the case in which $$A_{3}’ = 0$$.

17. Prove that if $$z^{3} + 3Hz + G = 0$$ has two complex roots then the equation $8\alpha^{3} + 6\alpha H – G = 0$ has one real root which is the real part $$\alpha$$ of the complex roots of the original equation; and show that $$\alpha$$ has the same sign as $$G$$.

18. An equation of any order with complex coefficients will in general have no real roots nor pairs of conjugate complex roots. How many conditions must be satisfied by the coefficients in order that the equation should have (a) a real root, (b) a pair of conjugate roots?

19. Coaxal circles. In Fig. 26, let $$a$$, $$b$$, $$z$$ be the arguments of $$A$$, $$B$$, $$P$$. Then $\operatorname{am}\frac{z – b}{z – a} = APB,$ if the principal value of the amplitude is chosen. If the two circles shown in the figure are equal, and $$z’$$, $$z_{1}$$, $$z_{1}’$$ are the arguments of $$P’$$, $$P_{1}$$, $$P_{1}’$$, and $$APB = \theta$$, it is easy to see that $\operatorname{am}\frac{z’ – b}{z’ – a} = \pi – \theta,\quad \operatorname{am}\frac{z_{1} – b}{z_{1} – a} = -\theta,$ and $\operatorname{am}\frac{z_{1}’ – b}{z_{1}’ – a} = -\pi + \theta.$

The locus defined by the equation $\operatorname{am}\frac{z – b}{z – a} = \theta,$ where $$\theta$$ is constant, is the arc $$APB$$. By writing $$\pi – \theta$$, $$-\theta$$, $$-\pi + \theta$$ for $$\theta$$, we obtain the other three arcs shown. The system of equations obtained by supposing that $$\theta$$ is a parameter, varying from $$-\pi$$ to $$+\pi$$, represents the system of circles which can be drawn through the points $$A$$, $$B$$. It should however be observed that each circle has to be divided into two parts to which correspond different values of $$\theta$$.

20. Now let us consider the equation $\begin{equation*}\left|\frac{z – b}{z – a}\right| = \lambda, \tag{1}\end{equation*}$ where $$\lambda$$ is a constant.

Let $$K$$ be the point in which the tangent to the circle $$ABP$$ at $$P$$ meets $$AB$$. Then the triangles $$KPA$$, $$KBP$$ are similar, and so $AP/PB = PK/BK = KA/KP = \lambda.$ Hence $$KA/KB = \lambda^{2}$$, and therefore $$K$$ is a fixed point for all positions of $$P$$ which satisfy the equation Eq(1). Also $$KP^{2} = KA \cdot KB$$, and so is constant. Hence the locus of $$P$$ is a circle whose centre is $$K$$.

The system of equations obtained by varying $$\lambda$$ represents a system of circles, and every circle of this system cuts at right angles every circle of the system of Ex. 19.

The system of Ex. 19 is called a system of coaxal circles of the common point kind. The system of Ex. 20 is called a system of coaxal circles of the limiting point kind, $$A$$ and $$B$$ being the limiting points of the system. If $$\lambda$$ is very large or very small then the circle is a very small circle containing $$A$$ or $$B$$ in its interior.

21. Bilinear Transformations. Consider the equation $\begin{equation*} z = Z + a, \tag{1} \end{equation*}$ where $$z = x + yi$$ and $$Z = X + Yi$$ are two complex variables which we may suppose to be represented in two planes $$xoy$$, $$XOY$$. To every value of $$z$$ corresponds one of $$Z$$, and conversely. If $$a = \alpha + \beta i$$ then $x = X + \alpha,\quad y = Y + \beta,$ and to the point $$(x, y)$$ corresponds the point $$(X, Y)$$. If $$(x, y)$$ describes a curve of any kind in its plane, $$(X, Y)$$ describes a curve in its plane. Thus to any figure in one plane corresponds a figure in the other. A passage of this kind from a figure in the plane $$xoy$$ to a figure in the plane $$XOY$$ by means of a relation such as (1) between $$z$$ and $$Z$$ is called a transformation. In this particular case the relation between corresponding figures is very easily defined. The $$(X, Y)$$ figure is the same in size, shape, and orientation as the $$(x, y)$$ figure, but is shifted a distance $$\alpha$$ to the left, and a distance $$\beta$$ downwards. Such a transformation is called a translation.

Now consider the equation $\begin{equation*} z = \rho Z, \tag{2} \end{equation*}$ where $$\rho$$ is real. This gives $$x = \rho X$$, $$y = \rho Y$$. The two figures are similar and similarly situated about their respective origins, but the scale of the $$(x, y)$$ figure is $$\rho$$ times that of the $$(X, Y)$$ figure. Such a transformation is called a magnification.

Finally consider the equation $\begin{equation*} z = (\cos\phi + i \sin\phi)Z. \tag{3} \end{equation*}$ It is clear that $$|z| = |Z|$$ and that one value of $$\operatorname{am} z$$ is $$\operatorname{am} Z + \phi$$, and that the two figures differ only in that the $$(x, y)$$ figure is the $$(X, Y)$$ figure turned about the origin through an angle $$\phi$$ in the positive direction. Such a transformation is called a rotation.

The general linear transformation $\begin{equation*} z = aZ + b \tag{4} \end{equation*}$ is a combination of the three transformations 1, (2), (3). For, if $$|a| = \rho$$ and $$\operatorname{am} a = \phi$$, we can replace (4) by the three equations $z = z’ + b,\quad z’ = \rho Z’,\quad Z’ = (\cos\phi + i\sin\phi)Z.$ Thus the general linear transformation is equivalent to the combination of a translation, a magnification, and a rotation.

Next let us consider the transformation $\begin{equation*} z = 1/Z. \tag{5} \end{equation*}$ If $$|Z| = R$$ and $$\operatorname{am} Z = \Theta$$, then $$|z| = 1/R$$ and $$\operatorname{am} z = -\Theta$$, and to pass from the $$(x, y)$$ figure to the $$(X, Y)$$ figure we invert the former with respect to $$o$$, with unit radius of inversion, and then construct the image of the new figure in the axis $$ox$$ (i.e. the symmetrical figure on the other side of $$ox$$).

Finally consider the transformation $\begin{equation*} z = \frac{aZ + b}{cZ + d}. \tag{6} \end{equation*}$ This is equivalent to the combination of the transformations $z = (a/c) + (bc – ad)(z’/c),\quad z’ = 1/Z’,\quad Z’ = cZ + d,$ i.e. to a certain combination of transformations of the types already considered.

The transformation (6) is called the general bilinear transformation. Solving for $$Z$$ we obtain $Z = \frac{dz – b}{cz – a}.$

The general bilinear transformation is the most general type of transformation for which one and only one value of $$z$$ corresponds to each value of $$Z$$, and conversely.

22. The general bilinear transformation transforms circles into circles. This may be proved in a variety of ways. We may assume the well-known theorem in pure geometry, that inversion transforms circles into circles (which may of course in particular cases be straight lines). Or we may use the results of Exs. 19 and 20. If, e.g., the $$(x, y)$$ circle is $|(z – \sigma)/(z – \rho)| = \lambda,$ and we substitute for $$z$$ in terms of $$Z$$, we obtain $|(Z – \sigma’)/(Z – \rho’)| = \lambda’,$ where $\sigma’ = -\frac{b – \sigma d}{a – \sigma c},\quad \rho’ = -\frac{b – \rho d}{a – \rho c},\quad \lambda’ = \left|\frac{a – \rho c}{a – \sigma c}\right|\lambda.$

23. Consider the transformations $$z = 1/Z$$, $$z = (1 + Z)/(1 – Z)$$, and draw the $$(X, Y)$$ curves which correspond to (1) circles whose centre is the origin, (2) straight lines through the origin.

24. The condition that the transformation $$z = (aZ + b)/(cZ + d)$$ should make the circle $$x^{2} + y^{2} = 1$$ correspond to a straight line in the $$(X, Y)$$ plane is $$|a| = |c|$$.

25. Cross ratios. The cross ratio $${(z_{1}, z_{2}; z_{3}, z_{4})}$$ is defined to be $\frac{(z_{1} – z_{3}) (z_{2} – z_{4})}{(z_{1} – z_{4}) (z_{2} – z_{3})}.$

If the four points $$z_{1}$$, $$z_{2}$$, $$z_{3}$$, $$z_{4}$$ are on the same line, this definition agrees with that adopted in elementary geometry. There are $$24$$ cross ratios which can be formed from $$z_{1}$$, $$z_{2}$$, $$z_{3}$$, $$z_{4}$$ by permuting the suffixes. These consist of six groups of four equal cross ratios. If one ratio is $$\lambda$$, then the six distinct cross ratios are $$\lambda$$, $$1 – \lambda$$, $$1/\lambda$$, $$1/(1 – \lambda)$$, $$(\lambda – 1)/\lambda$$, $$\lambda/(\lambda – 1)$$. The four points are said to be harmonic or harmonically related if any one of these is equal to $$-1$$. In this case the six ratios are $$-1$$, $$2$$, $$-1$$, $$\frac{1}{2}$$, $$2$$, $$\frac{1}{2}$$.

If any cross ratio is real then all are real and the four points lie on a circle. For in this case $\operatorname{am}\frac{(z_{1} – z_{3}) (z_{2} – z_{4})}{(z_{1} – z_{4}) (z_{2} – z_{3})}$ must have one of the three values $$-\pi$$, $$0$$, $$\pi$$, so that $$\operatorname{am}\{(z_{1} – z_{3})/(z_{1} – z_{4})\}$$ and $$\operatorname{am}\{(z_{2} – z_{3})/(z_{2} – z_{4})\}$$ must either be equal or differ by $$\pi$$ (cf. Ex. 19).

If $${(z_{1}, z_{2}; z_{3}, z_{4})} = – 1$$, we have the two equations $\operatorname{am}\frac{z_{1} – z_{3}}{z_{1} – z_{4}} = \pm\pi + \operatorname{am}\frac{z_{2} – z_{3}}{z_{2} – z_{4}},\quad \left|\frac{z_{1} – z_{3}}{z_{1} – z_{4}}\right| = \left|\frac{z_{2} – z_{3}}{z_{2} – z_{4}}\right|.$ The four points $$A_{1}$$, $$A_{2}$$, $$A_{3}$$, $$A_{4}$$ lie on a circle, $$A_{1}$$ and $$A_{2}$$ being separated by $$A_{3}$$ and $$A_{4}$$. Also $$A_{1}A_{3}/A_{1}A_{4} = A_{2}A_{3}/A_{2}A_{4}$$. Let $$O$$ be the middle point of $$A_{3}A_{4}$$. The equation $\frac{(z_{1} – z_{3}) (z_{2} – z_{4})}{(z_{1} – z_{4}) (z_{2} – z_{3})} = -1$ may be put in the form $(z_{1} + z_{2}) (z_{3} + z_{4}) = 2(z_{1}z_{2} + z_{3}z_{4}),$ or, what is the same thing, $\{z_{1} – \tfrac{1}{2}(z_{3} + z_{4})\} \{z_{2} – \tfrac{1}{2}(z_{3} + z_{4})\} = \{\tfrac{1}{2}(z_{3} – z_{4})\}^{2}.$ But this is equivalent to $$\overline{OA_{1}} \cdot \overline{OA_{2}} = \overline{OA_{3}}^{2} = \overline{OA_{4}}^{2}$$. Hence $$OA_{1}$$ and $$OA_{2}$$ make equal angles with $$A_{3}A_{4}$$, and $$OA_{1} \cdot OA_{2} = OA_{3}^{2} = OA_{4}^{2}$$. It will be observed that the relation between the pairs $$A_{1}$$, $$A_{2}$$ and $$A_{3}$$, $$A_{4}$$ is symmetrical. Hence, if $$O’$$ is the middle point of $$A_{1}A_{2}$$, $$O’A_{3}$$ and $$O’A_{4}$$ are equally inclined to $$A_{1}A_{2}$$, and $$O’A_{3} \cdot O’A_{4} = O’A_{1}^{2} = O’A_{2}^{2}$$.

26. If the points $$A_{1}$$, $$A_{2}$$ are given by $$az^{2} + 2bz + c = 0$$, and the points $$A_{3}$$, $$A_{4}$$ by $$a’z^{2} + 2b’z + c’ = 0$$, and $$O$$ is the middle point of $$A_{3}A_{4}$$, and $$ac’ + a’c – 2bb’ = 0$$, then $$OA_{1}$$, $$OA_{2}$$ are equally inclined to $$A_{3}A_{4}$$ and $$OA_{1} \cdot OA_{2} = OA_{3}^{2} = OA_{4}^{2}$$.

27. $$AB$$, $$CD$$ are two intersecting lines in Argand’s diagram, and $$P$$, $$Q$$ their middle points. Prove that, if $$AB$$ bisects the angle $$CPD$$ and $$PA^{2} = PB^{2} = PC \cdot PD$$, then $$CD$$ bisects the angle $$AQB$$ and $$QC^{2} = QD^{2} = QA \cdot QB$$.

28. The condition that four points should lie on a circle. A sufficient condition is that one (and therefore all) of the cross ratios should be real (Ex. 25); this condition is also necessary. Another form of the condition is that it should be possible to choose real numbers $$\alpha$$, $$\beta$$, $$\gamma$$ such that $\begin{vmatrix} 1 & 1 & 1\\ \alpha & \beta & \gamma\\ z_{1}z_{4} + z_{2}z_{3} & z_{2}z_{4} + z_{3}z_{1} & z_{3}z_{4} + z_{1}z_{2} \end{vmatrix} = 0.$

29. Prove the following analogue of De Moivre’s Theorem for real numbers: if $$\phi_{1}$$, $$\phi_{2}$$, $$\phi_{3}$$, … is a series of positive acute angles such that \begin{aligned} {2} \tan\phi_{m+1} &= \tan\phi_{m} \sec\phi_{1} &&+ \sec\phi_{m} \tan\phi_{1},\\ \tan\phi_{m+n} &= \tan\phi_{m} \sec\phi_{n} &&+ \sec\phi_{m} \tan\phi_{n},\\ \sec\phi_{m+n} &= \sec\phi_{m} \sec\phi_{n} &&+ \tan\phi_{m} \tan\phi_{n},\end{aligned} and $\tan\phi_{m} + \sec\phi_{m} = (\tan\phi_{1} + \sec\phi_{1})^{m}.$

[Use the method of mathematical induction.]

30. The transformation $$z = Z^{m}$$. In this case $$r = R^{m}$$, and $$\theta$$ and $$m\Theta$$ differ by a multiple of $$2\pi$$. If $$Z$$ describes a circle round the origin then $$z$$ describes a circle round the origin $$m$$ times.

The whole $$(x, y)$$ plane corresponds to any one of $$m$$ sectors in the $$(X, Y)$$ plane, each of angle $$2\pi/m$$. To each point in the $$(x, y)$$ plane correspond $$m$$ points in the $$(X, Y)$$ plane.

31. Complex functions of a real variable. If $$f(t)$$, $$\phi(t)$$ are two real functions of a real variable $$t$$ defined for a certain range of values of $$t$$, we call $\begin{equation*} z = f(t) + i\phi(t) \tag{1} \end{equation*}$ a complex function of $$t$$. We can represent it graphically by drawing the curve $x = f(t),\quad y = \phi(t);$ the equation of the curve may be obtained by eliminating $$t$$ between these equations. If $$z$$ is a polynomial in $$t$$, or rational function of $$t$$, with complex coefficients, we can express it in the form (1) and so determine the curve represented by the function.

(i) Let $z = a + (b – a)t,$ where $$a$$ and $$b$$ are complex numbers. If $$a = \alpha + \alpha’ i$$, $$b = \beta + \beta’ i$$, then $x = \alpha + (\beta – \alpha)t,\quad y = \alpha’ + (\beta’ – \alpha’)t.$ The curve is the straight line joining the points $$z = a$$ and $$z = b$$. The segment between the points corresponds to the range of values of $$t$$ from $$0$$ to $$1$$. Find the values of $$t$$ which correspond to the two produced segments of the line.

(ii) If $z = c + \rho\left(\frac{1 + ti}{1 – ti}\right),$ where $$\rho$$ is positive, then the curve is the circle of centre $$c$$ and radius $$\rho$$. As $$t$$ varies through all real values $$z$$ describes the circle once.

(iii) In general the equation $$z = (a + bt)/(c + dt)$$ represents a circle. This can be proved by calculating $$x$$ and $$y$$ and eliminating: but this process is rather cumbrous. A simpler method is obtained by using the result of Ex. 22. Let $$z = (a + bZ)/(c + dZ)$$, $$Z = t$$. As $$t$$ varies $$Z$$ describes a straight line, viz. the axis of $$X$$. Hence $$z$$ describes a circle.

(iv) The equation $z = a + 2bt + ct^{2}$ represents a parabola generally, a straight line if $$b/c$$ is real.

(v) The equation $$z = (a + 2bt + ct^{2})/(\alpha + 2\beta t + \gamma t^{2})$$, where $$\alpha$$, $$\beta$$, $$\gamma$$ are real, represents a conic section.

[Eliminate $$t$$ from $x = (A + 2Bt + Ct^{2})/(\alpha + 2\beta t + \gamma t^{2}),\quad y = (A’ + 2B’t + C’t^{2})/(\alpha + 2\beta t + \gamma t^{2}),$ where $$A + A’i = a$$, $$B + B’i = b$$, $$C + C’i = c$$.]

1. The numbers $$a + \sqrt{b}$$, $$a – \sqrt{b}$$, where $$a$$, $$b$$ are rational, are sometimes said to be‘conjugate’.↩︎
2. We suppose that as we go round the triangle in the direction $$ABC$$ we leave it on our left.↩︎