Let r and θ respectively be the modulus and amplitude of the complex number z = 2 - i 2 tan 5 π 8 , then ( r…

Given: $z=2-i\left(2\tan \frac{5\pi }{8}\right)$

$\Rightarrow z=2\left[1-i\tan \left(\frac{5\pi }{8}\right)\right]$

$\Rightarrow z=2\left[1-\frac{i\sin \left({\displaystyle \frac{5\pi }{8}}\right)}{\cos \left(\frac{5\pi }{8}\right)}\right]$

$\Rightarrow z=\frac{2}{\cos \left(\frac{5\pi }{8}\right)}\left[\cos \left(\frac{5\pi }{8}\right)-i\sin \left(\frac{5\pi }{8}\right)\right]$

$\Rightarrow z=\frac{2}{\cos \left(\pi -\frac{3\pi }{8}\right)}\left[\cos \left(\pi -\frac{3\pi }{8}\right)-i\sin \left(\pi -\frac{3\pi }{8}\right)\right]$

$\Rightarrow z=\frac{2}{-\cos \left(\frac{3\pi }{8}\right)}\left[-\cos \left(\frac{3\pi }{8}\right)-i\sin \left(\frac{3\pi }{8}\right)\right]$

$\Rightarrow z=\frac{2}{\cos \left(\frac{3\pi }{8}\right)}\left[\cos \left(\frac{3\pi }{8}\right)+i\sin \left(\frac{3\pi }{8}\right)\right]$

$\Rightarrow z=2\sec \frac{3\pi }{8}{e}^{i·\frac{3\pi }{8}}$

Now on comparing with $z=\left|z\right|e^{i\theta }$ we get,

$\Rightarrow \theta =\frac{3\pi }{8}, r=2\sec \frac{3\pi }{8}$