====== 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. =====Question 7(i)===== 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. =====Question 7(ii)===== 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*} =====Question 7(iii)===== 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. =====Question 7(iv)===== 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. =====Question 7(v)===== 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. =====Question 7(vi)===== 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. ====Go to ==== [[math-11-nbf:sol:unit01:ex1-4-p7|< Question 6(x-xvii)]] [[math-11-nbf:sol:unit01:ex1-4-p9|Question 8 >]]