Question 2, Exercise 1.4

Solutions of Question 2 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 the complex number $\left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\left(\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}\right)$ in rectangular form.

Solution.

Let $z_1=\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}=e^{i\frac{\pi}{6}}$ and $z_2=\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}=e^{i\frac{\pi}{3}}$. Then $$\begin{align} z_1 z_2 & = e^{i\frac{\pi}{6}} \cdot e^{i\frac{\pi}{3}} \\ & = e^{i\left(\frac{\pi}{6}+\frac{\pi}{3}\right)} \\ & = e^{i\frac{\pi}{2}} \\ & = \cos \dfrac{\pi}{2} +i \sin \dfrac{\pi}{2} \\ & = 0 + i (1) = i. \end{align}$$ Hence, we proved $$\left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\left(\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}\right) = i.$$

Alternative Method:

$$\begin{align} &\left(\cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6}\right)\left(\cos \dfrac{\pi}{3} + i \sin \dfrac{\pi}{3}\right)\\ =& \left(\cos \dfrac{\pi}{6} \cos \dfrac{\pi}{3} - \sin \dfrac{\pi}{6} \sin \dfrac{\pi}{3}\right) + i \left(\sin \dfrac{\pi}{6} \cos \dfrac{\pi}{3} + \cos \dfrac{\pi}{6} \sin \dfrac{\pi}{3}\right) \\ =& \left(\dfrac{\sqrt{3}}{2} \cdot \dfrac{1}{2} - \dfrac{1}{2} \cdot \dfrac{\sqrt{3}}{2}\right) + i \left(\dfrac{1}{2} \cdot \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{3}}{2}\right) \\ =& \left(\dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4}\right) + i \left(\dfrac{1}{4} + \dfrac{3}{4}\right) \\ =& 0 + i \cdot 1 \\ =& i \end{align}$$ Which is rectangular form.

Write the complex number $\dfrac{\cos \dfrac{\pi}{6} - i \sin \dfrac{\pi}{6}}{2\left(\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}\right)}$ in rectangular form.

Solution.

$$\begin{align} &\dfrac{\cos \dfrac{\pi}{6} - i \sin \dfrac{\pi}{6}}{2\left(\cos \dfrac{\pi}{3} + i \sin \dfrac{\pi}{3}\right)}\\ =&\dfrac{\cos (-\dfrac{\pi}{6}) + i \sin (-\dfrac{\pi}{6})}{2\left(\cos \dfrac{\pi}{3} + i \sin \dfrac{\pi}{3}\right)}\\ =& \dfrac{1}{2} \left[\cos \left(-\dfrac{\pi}{6} -\dfrac{\pi}{3}\right) + i \sin \left(-\dfrac{\pi}{6}- \dfrac{\pi}{3}\right)\right] \\ =& \dfrac{1}{2} \left[\cos \left(-\dfrac{3\pi}{6}\right) + i \sin \left(\dfrac{-3\pi}{6}\right)\right] \\ =& \dfrac{1}{2} \left[\cos \left(\dfrac{-\pi}{2}\right) + i \sin \left(\dfrac{-\pi}{2}\right)\right] \\ =& \dfrac{1}{2} \left[0 - i (1)\right] \\ =& -\dfrac{1}{2}i \end{align}$$ Which is rectangular form.

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