Question

Complex number z = $\frac{i-1}{cos(\pi /3)+isin(\pi /3)}$ in polar form is

• A. $r=\sqrt{2}(cos\frac{5\pi }{12}+isin\frac{5\pi }{12})$
• B. $r=\sqrt{2}(cos\frac{\pi }{4}+isin\frac{\pi }{4})$
• C. $r=\sqrt{2}(cos\frac{\pi }{6}+isin\frac{\pi }{6})$
• D. $Noneoftheabove$

Answer 1

Answer: (A)

$z=\frac{i-1}{cos\left(\frac{\pi }{3}\right)+isin\left(\frac{\pi }{3}\right)}$
$=\frac{i-1}{\left(\frac{1+i\sqrt{3}}{2}\right)}=\frac{2\left(i-1\right)\left(1-i\sqrt{3}\right)}{\left(1-i^2\cdot 3\right)}$
$=\frac{2}{4}\left(i-1-i^2\sqrt{3}+i\sqrt{3}\right)$
$=\frac{1}{2}\left(\sqrt{3}-1+i\sqrt{3}+i\right)$
$=\frac{\left(\sqrt{3}-1\right)}{2}+\frac{i\left(\sqrt{3}+1\right)}{2}\dots \left(i\right)$
Let $rcos\theta =\frac{\sqrt{3}-1}{2}\dots \left(ii\right)$
$rsin\theta =\frac{\sqrt{3}+1}{2}\dots \left(iii\right)$
On squaring and adding Eqs. $\left(ii\right)$ and $\left(iii\right)$
$r^2=\frac{1}{4}\left\{3+1-2\sqrt{3}+3+1+2\sqrt{3}\right\}$
$r^2=2$
$\Rightarrow r=\sqrt{2}$
From on dividing Eq. $\left(iii\right)$ from Eq. $\left(ii\right)$
$tan\theta =\frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$tan\theta =\frac{{\left(\sqrt{3}+1\right)}^2}{3-1}=\frac{3+1+2\sqrt{3}}{2}$
$tan\theta =2+\sqrt{3}$
$\theta =\frac{7\pi }{12}$
Principal value of $\theta =\pi -\frac{7\pi }{12}$
$=\frac{5\pi }{12}$
Hence, the polar form is
$r=\sqrt{2}\left(cos\frac{5\pi }{12}+isin\frac{5\pi }{12}\right)$