Q73 / 180 · COMEDK 2017 · Mathematics

If $(\sqrt{5}+\sqrt{3} i)^{33}=2^{49} z$, then modulus of the complex number $z$ is equal to

  • A
    1
  • Correct answer
    B
    $\sqrt{2}$
  • C
    $2 \sqrt{2}$
  • D
    4

Solution

We have, $(\sqrt{5}+\sqrt{3} i)^{33}=2^{49} z$ Taking modulus on both sides, we get $\begin{aligned} &\left.2^{49}|z|=\mid \sqrt{5}+\sqrt{3} i\right)^{33} \\ \Rightarrow & 2^{49}|z|=(\sqrt{5+3})^{33}=(\sqrt{8})^{33}=2^{33} 2^{33 / 2} \\ \Rightarrow &|z|=\frac{2^{33} 2^{33 / 2}}{2^{49}}=2^{\frac{33}{2}-16}=2^{1 / 2}=\sqrt{2} \end{aligned}$

Hence, the correct answer is option: 'B'.

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