Question

Convert the complex number $z=\frac{i-1}{\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}}$ in the polar form.

Answer 1

Given data: $z=\frac{i-1}{\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}}$
$z=\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2}i}\left\{\right(\because \cos \frac{\pi }{3}=\frac{1}{2}\text{and}\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$}
$z=\frac{2(i-1)}{1+\sqrt{3}i}$
Step 2 Do rationalisation
$z=\frac{2(i-1)}{1+\sqrt{3i}}\times \frac{1-\sqrt{3}i}{1-\sqrt{3}i}$
$Z=\frac{2(i+\sqrt{3}-1+\sqrt{3}i)}{(1{)}^2-(\sqrt{3}i{)}^2}$
$z=\frac{2\left[\right(\sqrt{3}-1)+(\sqrt{3}+1\left)i\right]}{1+3}\Rightarrow z=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2}i$
Compare the form of a complex number
$x+iy=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2}i$
So, $x=\frac{\sqrt{3}-1}{2}$ and $y=\frac{\sqrt{3}+1}{2}$
$\therefore P=(\frac{\sqrt{3}-1}{2},\frac{\sqrt{3}+1}{2})$
Step 4 Find out the polar coordinates of the point.
Let $x=r\cos \theta$ and $y=r\sin \theta$
$\therefore \frac{\sqrt{3}-1}{2}=r\cos \theta$ and $\frac{\sqrt{3}+1}{2}=r\sin \theta$
$\frac{\sqrt{3}-1}{2r}=\cos \theta$ and $\frac{\sqrt{3}+1}{2r}=\sin \theta$
Now, $r=\sqrt{x^2+y^2}=\sqrt{{\left(\frac{\sqrt{3}-1}{2}\right)}^2+{\left(\frac{\sqrt{3}+1}{2}\right)}^2}$
$=\frac{1}{2}\sqrt{(3+1-2\sqrt{3})+(3+1+2\sqrt{3})}=\frac{2\sqrt{2}}{2}=\sqrt{2}$
$\therefore r=\sqrt{2}$ which gives cos $\theta =\frac{\sqrt{3}-1}{2\sqrt{2}}$ and $\sin \theta =\frac{\sqrt{3}+1}{2\sqrt{2}}$
$\therefore \theta =\frac{\pi }{4}+\frac{\pi }{6}\Rightarrow \theta =\frac{5\pi }{12}$
Step 5 Represent the polar form
$z=r(\cos \theta +i\sin \theta )$
$z=\sqrt{2}(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12})$