Question

The modulus-amplitude form of $\frac{(1-i)^{3}(2-i)}{(2+i)(1+i)}$ is

• A. $2 \text{cis}\left(\pi-\tan ^{-1} \frac{4}{3}\right)$
• B. $2 \text{cis}\left(-\tan ^{-1} \frac{4}{3}\right)$
• C. $2 \text{cis}\left(-\pi+\tan ^{-1} \frac{4}{3}\right)$
• D. $2 \text{cis}\left(\tan ^{-1} \frac{4}{3}\right)$

Answer 1

Answer: (A)

We have, $\frac{(1-i)^{3}(2-i)}{(2+i)(1+i)}=\frac{(1-i)^{2}(1-i)(2-i)}{(1+i)(2+i)}$
$=\frac{-2 i(1-i)(1-i)(2-i)(2-i)}{(1+i)(1-i)(2+i)(2-i)}$
$=\frac{-2 i(-2 i)(3-4 i)}{2 \cdot 5}=-2\left(\frac{3}{5}-\frac{4 i}{5}\right)$
$=2\left(\frac{-3}{5}+\frac{4}{5} i\right)$
$=2 cis\left(\pi-\tan ^{-1}\left(\frac{4}{3}\right)\right)$