Mathematics | TS EAMCET 2018 | Discussion Page

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

$2cis\left( \pi-\tan^{-1}\frac{4}{3}\right)$

$2cis\left( -\tan^{-1}\frac{4}{3}\right)$

$2cis\left( -\pi+\tan^{-1}\frac{4}{3}\right)$

$2cis\left( \tan^{-1}\frac{4}{3}\right)$

Option A

We have,

$\frac{(1-i)^{3}(2-i)}{(2+i)(1+i)}=\frac{(1-i)^{2}(1-i)(2-i)}{(1+i)(2+i)}$

$=\frac{-2i(1-i)^{}(1-i)(2-i)(2-i)}{(1+i)(1-i)(2+i)(2-i)}$

$=\frac{-2i(-2i)(3-4i)}{2.5}=-2\left(\frac{3}{5}-\frac{4i}{5}\right)$

$=-2\left(\frac{-3}{5}+\frac{4}{5}i\right)$

$2cis\left( \pi-\tan^{-1}\frac{4}{3}\right)$

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