Explanation

(b)
Given, $\left( \sqrt{5} + \sqrt{3}i \right)^{33} = 2^{49}z$
Let $\sqrt{5} = r\cos\theta,\ \ \ \ \sqrt{3} = r\sin\theta$
$\therefore\ \ \ \ \ r^{2} = 5 + 3\ \ \ \ \Rightarrow \ \ \ \ r = 2\sqrt{2}$
$\therefore\ \ \ \left( r\cos{\theta + ir\sin\theta} \right)^{33} = 2^{49}z$
$\Rightarrow \ \ \ \ \left| r^{33}\left( \cos{33\theta} + i\sin{33\theta} \right) \right| = |2^{49}z|$
$\Rightarrow \ \ \ \ \left( 2\sqrt{2} \right)^{33}|\cos{33\theta + i\sin{33\theta\left| = 2^{49} \right|z|}}$
$\Rightarrow \ \ \ \ \ 2^{\frac{99}{2}}(1) = 2^{49}|z|\ \ \ \ \ \Rightarrow \ \ \ \ \ |z| = \sqrt{2}$