Question 7, Exercise 1.4

Solutions of Question 7 of Exercise 1.4 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Convert the following equation in Cartesian form: $\arg (z-1)=-\dfrac{\pi}{4}$

Solution.

Suppose $z=x+iy$, as $$\begin{align*} &\arg (z-1)=-\dfrac{\pi}{4} \\ \implies & \arg(x+iy-1) = -\dfrac{\pi}{4} \\ \implies & \arg(x-1+iy) = -\dfrac{\pi}{4} \\ \implies & \tan^{-1}\left(\dfrac{y}{x-1}\right) = -\dfrac{\pi}{4} \\ \implies & \dfrac{y}{x-1} = \tan\left(-\dfrac{\pi}{4}\right) \\ \implies & \dfrac{y}{x-1} = -1 \\ \implies & y = -x+1 \\ \implies & x+y = 1. \end{align*}$$ As required.

Convert the following equations and inequations in Cartesian form: $z \bar{z}=4\left|e^{i \theta}\right|$

Solution.

Suppose $z=x+iy$, then $\bar{z}=x-iy$. As $$\begin{align*} &z \bar{z}=4\left|e^{i \theta}\right| \\ \implies & (x+iy)(x-iy)=4|\cos\theta+i\sin\theta| \\ \implies & x^2+y^2 = 4\sqrt{\cos^2\theta+\sin^2\theta} \\ \implies & x^2+y^2 = 4\sqrt{1} \\ \implies & x^2+y^2 = 4. \end{align*}$$

Convert the following equations and inequations in Cartesian form: $-\dfrac{\pi}{3} \leq \arg (z-4) \leq \dfrac{\pi}{3}$

Solution.

$$\begin{align*} &-\frac{\pi}{3} \leq \arg (z-4) \leq \frac{\pi}{3}\\ \implies & -\frac{\pi}{3} \leq \arg(x+iy-4) \leq \frac{\pi}{3} \\ \implies & -\frac{\pi}{3} \leq \arg(x-4+iy) \leq \frac{\pi}{3} \\ \implies & -\frac{\pi}{3} \leq \tan^{-1}\left(\frac{y}{x-4}\right) \leq \frac{\pi}{3} \\ \implies & \tan\left(-\frac{\pi}{3}\right) \leq \frac{y}{x-4} \leq \tan\left(\frac{\pi}{3}\right) \\ \implies & -\sqrt{3}\leq \frac{y}{x-4} \leq \sqrt{3}. \end{align*}$$ As required.

Convert the following equations and inequations in Cartesian form: $0 \leq \arg \left(\dfrac{z-4}{1+i}\right) \leq \dfrac{\pi}{6}$

Solution.

$$\begin{align*} &0 \leq \arg \left(\frac{z-4}{1+i}\right) \leq \frac{\pi}{6} \\ \implies & 0 \leq \arg\left(\frac{x+iy-4}{1+i}\right) \leq \frac{\pi}{6} \\ \implies & 0 \leq \arg\left(\frac{x-4+iy}{1+i}\right) \leq \frac{\pi}{6} \\ \implies & 0 \leq \arg\left(\frac{x-4+iy}{1+i}\times \frac{1-i}{1-i}\right) \leq \frac{\pi}{6} \\ \implies & 0 \leq \arg\left(\frac{x-4+y-i(x-4)+iy}{1+1}\right) \leq \frac{\pi}{6} \\ \implies & 0 \leq \arg\left(\frac{x+y-4+i(y-x+4)}{2}\right) \leq \frac{\pi}{6} \\ \implies & 0 \leq \arg\left(\frac{x+y-4}{2}+i\frac{y-x+4}{2}\right) \leq \frac{\pi}{6} \\ \implies & 0 \leq \tan^{-1}\left(\frac{y-x+4}{x+y-4}\right) \leq \frac{\pi}{6} \\ \implies & \tan(0) \leq \frac{y-x+4}{x+y-4} \leq \tan\left(\frac{\pi}{6}\right) \\ \implies & 0 \leq \frac{y-x+4}{x+y-4} \leq \frac{1}{\sqrt{3}} \\ \implies & 0 \leq \sqrt{3}\left(\frac{y-x+4}{x+y-4}\right) \leq 1. \end{align*}$$ As required.

Convert the following equations and inequations in Cartesian form: $\arg \left(\dfrac{1-iz}{1-z}\right)=\dfrac{\pi}{4} ; z \neq i$

Solution.

Given $$\begin{align*} \arg \left(\dfrac{1-iz}{1-z}\right)=\dfrac{\pi}{4} ... (1) \end{align*}$$ Using the identity $$\arg\left(\dfrac{\theta_1}{\theta_2}\right)=\arg(\theta_1)-\arg(\theta_2),$$ we have $$\begin{align*} &\arg \left(\dfrac{1-iz}{1-z}\right) \\ = & \arg(1-iz) - \arg(1-z) \\ = & \arg(1-i(x+iy))-\arg(1-(x+iy)) \quad \text{ as } z=x+iy \\ = & \arg(1-ix-i^2y)-\arg(1-x-iy) \\ = & \arg(1+y-ix)-\arg(1-x-iy) \\ = & \tan^{-1}\left(\dfrac{-x}{1+y}\right) - \tan^{-1}\left(\dfrac{-y}{1-x}\right) \\ = & -\tan^{-1}\left(\dfrac{x}{1+y}\right) + \tan^{-1}\left(\dfrac{y}{1-x}\right) \quad \because \,\, \tan^{-1}(-\theta)=-\tan^{-1}(\theta) \\ = & \tan^{-1}\left(\dfrac{y}{1-x}\right) -\tan^{-1}\left(\dfrac{x}{1+y}\right)\\ \end{align*}$$ Using it in $(1)$, we get $$\begin{align*} \tan^{-1}\left(\dfrac{y}{1-x}\right) -\tan^{-1}\left(\dfrac{x}{1+y}\right)=\dfrac{\pi}{4} \end{align*}$$ Now using the identity, $$\begin{align*} \tan^{-1}A+\tan^{-1}B=\tan^{-1}\left(\frac{A+B}{1-AB}\right), \end{align*}$$ we have $$\begin{align*} & \tan^{-1} \left( \frac{\frac{y}{1-x} - \frac{x}{1+y}}{1 + \frac{y}{1-x} \cdot \frac{x}{1+y}} \right) = \frac{\pi}{4} \\ \implies & \frac{\frac{y + y^2 - x + x^2}{(1-x)(1+y)}}{\frac{(1-x)(1+y)+xy}{(1-x)(1+y)}} = \tan \frac{\pi}{4} \\ \implies & \frac{x^2+y^2 - x + y}{1-x+y-xy+xy} = 1 \\ \implies & x^2+y^2 - x + y=1-x+y \\ \implies & x^2+y^2=1. \end{align*}$$ As required.

Convert the following equations and inequations in Cartesian form: $\dfrac{1}{2} \arg (z-i)=\dfrac{\pi}{3}-\dfrac{1}{2} \arg (z+i)$

Solution.

Assume $z=x+iy$, then $$\begin{align*} &\dfrac{1}{2} \arg (z-i)=\dfrac{\pi}{3}-\dfrac{1}{2} \arg (z+i) \\ \implies & \dfrac{1}{2} \arg (z-i)+\dfrac{1}{2} \arg (z+i)=\dfrac{\pi}{3} \\ \implies & \arg (x+iy-i)+\arg (x+iy+i)=\dfrac{2\pi}{3} \\ \implies & \arg (x+iy-i)+\arg (x+iy+i)=\dfrac{2\pi}{3} \\ \implies & \arg (x+i(y-1))+\arg (x+i(y+1))=\dfrac{2\pi}{3} \\ \implies & \tan^{-1}\left(\dfrac{y-1}{x}\right)+\tan^{-1}\left(\dfrac{y+1}{x}\right)=\dfrac{2\pi}{3}. \end{align*}$$ Now using the identity, $$\begin{align*} \tan^{-1}A+\tan^{-1}B=\tan^{-1}\left(\frac{A+B}{1-AB}\right), \end{align*}$$ we have $$\begin{align*} &\tan^{-1}\left(\dfrac{\frac{y-1}{x}+\frac{y+1}{x}}{1-\left(\frac{y-1}{x}\right)\left(\frac{y+1}{x}\right)} \right)=\dfrac{2\pi}{3} \\ \implies & \dfrac{\frac{y-1+y+1}{x}}{1-\frac{y^2-1}{x^2}}=\tan\left(\dfrac{2\pi}{3}\right) \\ \implies & \dfrac{\frac{2y}{x}}{\frac{x^2-y^2+1}{x^2}}=-\sqrt{3} \\ \implies & \frac{2xy}{x^2-y^2+1}=-\sqrt{3} \\ \implies & 2xy=-\sqrt{3}(x^2-y^2+1) \\ \implies & \sqrt{3}(x^2-y^2+1)+2xy=0. \end{align*}$$ As required.

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