Question 1, Exercise 1.4

Solutions of Question 1 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.

Write a complex number $2+i 2 \sqrt{3}$ in polar form.

Solution.

Let $z=x+iy=2 + i 2 \sqrt{3}$. We have $$\begin{align} r & = \sqrt{x^2 + y^2} = \sqrt{2^2 + (2\sqrt{3})^2} \\ & = \sqrt{4 + 12} = \sqrt{16} = 4. \end{align}$$ and $$\begin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{2\sqrt{3}}{2}\right|\\ & = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}. \end{align}$$

Since the complex number $2 + i 2 \sqrt{3}$ lies in the first quadrant, the principal value of the argument $\theta$ is: $$\theta = \alpha = \frac{\pi}{3}.$$

Hence $$2 + i 2 \sqrt{3} = 4 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right).$$

Write the following complex number $3-i \sqrt{3}$ in polar form.

Solution.

Let $z=x+iy=3-i \sqrt{3}$ $$\begin{align} r &= \sqrt{x^2 + y^2} = \sqrt{3^2 + (-\sqrt{3})^2}\\ &= \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}. \end{align}$$

Next, $$\begin{align} \alpha &= \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{-\sqrt{3}}{3}\right| \\ &= \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ &= \frac{\pi}{6}. \end{align}$$

Since the complex number $3 - i \sqrt{3}$ lies in the fourth quadrant, the principal value of the argument $\theta$ is: $$\theta = -\alpha = -\frac{\pi}{6}.$$

Hence $$\begin{align} 3 - i \sqrt{3} & = 2\sqrt{3} \left( \cos\left( -\frac{\pi}{6} \right) + i \sin\left( -\frac{\pi}{6} \right) \right) \\ & = 2\sqrt{3} \left( \cos\left( \frac{\pi}{6} \right) - i \sin\left( \frac{\pi}{6} \right) \right). \end{align}$$

Write the following complex number $-2-i 2$ in polar form.

Solution.

Let $z=x+iy=-2-2i$ $$\begin{align} r &= \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (-2)^2} \\ &= \sqrt{4 + 4} = \sqrt{8} \\ &= 2\sqrt{2}. \end{align}$$

Next $$\begin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\right| \\ &= \tan^{-1}\left|\frac{-2}{-2}\right| \\ &= \tan^{-1}(1) = \frac{\pi}{4}. \end{align}$$ Since the complex number $-2 - i 2$ lies in the third quadrant, the principal value of the argument $\theta$ is: $$\begin{align} \theta = \pi + \alpha = \pi + \frac{\pi}{4} = \frac{5\pi}{4}. \end{align}$$ Hence $$\begin{align} -2 - i 2 = 2\sqrt{2} \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right). \end{align}$$

Write the following complex number $\dfrac{i-1}{\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}}$ in polar form.

Solution.

Let $z=x+iy= i - 1=-1+i$. Then $$\begin{align} r & = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (1)^2}\\ &= \sqrt{1 + 1} = \sqrt{2}. \end{align}$$ Next, $$\begin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\right| \\ &= \tan^{-1}\left|\frac{1}{-1}\right| \\ &= \tan^{-1}(1) = \frac{\pi}{4}. \end{align}$$

Since the complex number $i - 1$ lies in the second quadrant, the principal value of the argument $\theta$ is: $$\begin{align} \theta = \pi - \alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}. \end{align}$$

Therefore, the polar form of the complex number $i - 1$ is: $$\begin{align} i - 1 = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right). \end{align}$$ Now $$\begin{align} &\dfrac{i-1}{\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}} \\ =& \dfrac{\sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)}{\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}} \\ =&\sqrt{2} \left( \cos \left(\frac{3\pi}{4}- \frac{\pi}{3}\right)+ i \sin \left(\frac{3\pi}{4}- \frac{\pi}{3}\right) \right)\\ =&\sqrt{2} \left( \cos \frac{5\pi}{12}+ i \sin \frac{5\pi}{12} \right). \end{align}$$

Question 2>