Question
Write the complex number $\text{z}=\frac{1-\text{i}}{\cos\frac{\pi}{3}+\text{i}\sin\frac{\pi}{3}}$ in polar form.
✓
Answer
Given that,
$\text{z}=\frac{1-\text{i}}{\frac{1}{2}+\text{i}\frac{\sqrt{3}}{2}}=\frac{2-2\text{i}}{1+\text{i}\sqrt{3}}$
$=\frac{2-2\text{i}}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$
$\Rightarrow\text{z}=\frac{2-2\sqrt{3}\text{i}-2\text{i}+2\sqrt{3}\text{i}^2}{(1)^2-(\text{i}\sqrt{3})^2}=\frac{2-2\sqrt{3}\text{i}-2\text{i}-2\sqrt{3}}{1-3\text{i}^2}$
$=\frac{(2-2\sqrt{3})-(2+2\sqrt{3})\text{i}}{4}=\frac{1-\sqrt{3}}{2}-\frac{1+\sqrt{3}}{2}\text{i}$
$\Rightarrow\text{r}=\sqrt{\Big(\frac{1-\sqrt{3}}{2}\Big)^2+\Big(-\frac{1+\sqrt{3}}{2}\Big)^2}$
$=\sqrt{\frac{1+3-2\sqrt{3}}{4}+\frac{1+3+2\sqrt{3}}{4}}$
$=\sqrt{\frac{4-2\sqrt{3}+4+2\sqrt{3}}{4}}=\sqrt{\frac{8}{4}}$
So, $\text{r}=\sqrt{2}$
Now, $\text{arg(z)}=\tan^{-1}\frac{\text{y}}{\text{x}}$
$\Rightarrow\ \theta=\tan^{-}\frac{\Big(\frac{1+\sqrt{3}}{2}\Big)}{\Big(\frac{1-\sqrt{3}}{2}\Big)}=\tan^{-1}\Big[-\Big(\frac{1+\sqrt{3}}{1-\sqrt{3}}\Big)\Big]$
$\Rightarrow\ \theta=\tan^{-1}\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$\Rightarrow\ \theta=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}+\frac{\pi}{6}\Big)\Big]$ $\Bigg[\because\ \tan\Big(\frac{\pi}{4}+\frac{\pi}{6}\Big)=\frac{\tan\frac{\pi}{4}+\tan\frac{\pi}{6}}{1-\tan\frac{\pi}{4}\tan\frac{\pi}{6}}\Bigg]$
$\Rightarrow\ \theta=\frac{5\pi}{12}$
Hence, the polar is
$\text{z}=\sqrt{2}\Big[\cos\Big(\frac{5\pi}{12}\Big)+\text{i}\sin\Big(\frac{5\pi}{12}\Big)\Big]$
$\text{z}=\frac{1-\text{i}}{\frac{1}{2}+\text{i}\frac{\sqrt{3}}{2}}=\frac{2-2\text{i}}{1+\text{i}\sqrt{3}}$
$=\frac{2-2\text{i}}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$
$\Rightarrow\text{z}=\frac{2-2\sqrt{3}\text{i}-2\text{i}+2\sqrt{3}\text{i}^2}{(1)^2-(\text{i}\sqrt{3})^2}=\frac{2-2\sqrt{3}\text{i}-2\text{i}-2\sqrt{3}}{1-3\text{i}^2}$
$=\frac{(2-2\sqrt{3})-(2+2\sqrt{3})\text{i}}{4}=\frac{1-\sqrt{3}}{2}-\frac{1+\sqrt{3}}{2}\text{i}$
$\Rightarrow\text{r}=\sqrt{\Big(\frac{1-\sqrt{3}}{2}\Big)^2+\Big(-\frac{1+\sqrt{3}}{2}\Big)^2}$
$=\sqrt{\frac{1+3-2\sqrt{3}}{4}+\frac{1+3+2\sqrt{3}}{4}}$
$=\sqrt{\frac{4-2\sqrt{3}+4+2\sqrt{3}}{4}}=\sqrt{\frac{8}{4}}$
So, $\text{r}=\sqrt{2}$
Now, $\text{arg(z)}=\tan^{-1}\frac{\text{y}}{\text{x}}$
$\Rightarrow\ \theta=\tan^{-}\frac{\Big(\frac{1+\sqrt{3}}{2}\Big)}{\Big(\frac{1-\sqrt{3}}{2}\Big)}=\tan^{-1}\Big[-\Big(\frac{1+\sqrt{3}}{1-\sqrt{3}}\Big)\Big]$
$\Rightarrow\ \theta=\tan^{-1}\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$\Rightarrow\ \theta=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}+\frac{\pi}{6}\Big)\Big]$ $\Bigg[\because\ \tan\Big(\frac{\pi}{4}+\frac{\pi}{6}\Big)=\frac{\tan\frac{\pi}{4}+\tan\frac{\pi}{6}}{1-\tan\frac{\pi}{4}\tan\frac{\pi}{6}}\Bigg]$
$\Rightarrow\ \theta=\frac{5\pi}{12}$
Hence, the polar is
$\text{z}=\sqrt{2}\Big[\cos\Big(\frac{5\pi}{12}\Big)+\text{i}\sin\Big(\frac{5\pi}{12}\Big)\Big]$
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