Question

If (root 5 + root 3 i)^33=2^49 z , then modules of the complex number z is equal to . Please give answer with steps

Answer 1

SOLUTION :

GIVEN

$\sf{ {( \sqrt{5} + \sqrt{3}i) }^{33} = {2}^{49} \: z \: }$

TO DETERMINE

The modules of the complex number z i.e | z |

CONCEPT TO BE IMPLEMENTED

PROPERTIES OF COMPLEX NUMBERS

$\sf{1. \: \: \: if \: z = a +ib \: \: then \: \: |z| = \sqrt{ {a}^{2} + {b}^{2} } \: }$

$\sf{2. \: \: | {z}^{n} | = { |z| }^{n} }$

EVALUATION

$\sf{ {( \sqrt{5} + \sqrt{3}i) }^{33} = {2}^{49} \: z \: }$

Taking modulus in both sides we get

$\implies \sf{ | {( \sqrt{5} + \sqrt{3}i) }^{33} | = | {2}^{49} \: z| \: }$

$\implies \sf{ { |( \sqrt{5} + \sqrt{3}i)| }^{33} = {2}^{49} |z| \: }$

$\implies \sf{ { \bigg( \sqrt{ {( \sqrt{5} )}^{2} + {( \sqrt{3}) }^{2} } \bigg) }^{33} = {2}^{49} |z| \: }$

$\implies \sf{ { \bigg( \sqrt{8} \bigg) }^{33} = {2}^{49} |z| \: }$

$\displaystyle \implies \sf{ { \bigg( 8\bigg) }^{ \frac{33}{2} } = {2}^{49} |z| \: }$

$\displaystyle \implies \sf{ { \bigg( {2}^{3} \bigg) }^{ \frac{33}{2} } = {2}^{49} |z| \: }$

$\displaystyle \implies \sf{ { ( {2} )}^{ \frac{99}{2} } = {2}^{49} |z| \: }$

$\displaystyle \implies \sf{ |z| = {2}^{ (\frac{99}{2} - 49)} \: }$

$\displaystyle \implies \sf{ |z| = {2}^{ \frac{(99 - 98)}{2} } \: }$

$\implies \displaystyle \sf{ |z| = {2}^{ \frac{1}{2} } \: }$

$\implies \displaystyle \sf{ |z| = \sqrt{2} } \:$

RESULT

$\boxed{ \displaystyle \sf{ \: \: |z| = \sqrt{2} } \: \: \: }$

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