If $ {\left( {\sqrt 5 + \sqrt {3i} } \right)^{33}} = {2^{49}}z $ , then modulus of the complex number is
A.1
B. $ \sqrt 2 $
C. $ 2\sqrt 2 $
D.4

Answer
VerifiedVerified
137.1k+ views
Hint: The complex number in the question is z so we have to find modulus of z which is $ \left| z \right| $ . For finding this we have to take modulus on both sides of the equation. To find the solution properties of complex numbers are to be used.

Complete step-by-step answer:
According to the question,
Given: $ {\left( {\sqrt 5 + \sqrt {3i} } \right)^{33}} = {2^{49}}z $
Now, we have to find modulus of z which is $ \left| z \right| $ , so
By taking modulus on both sides we get,
 $ \left| {{{\left( {\sqrt 5 + \sqrt {3i} } \right)}^{33}}} \right| = \left| {{2^{49}}z} \right| $
By using the property of complex number $ \left| {{z^n}} \right| = {\left| z \right|^n} $ , we get
 $ {\left| {\left( {\sqrt 5 + \sqrt {3i} } \right)} \right|^{33}} = {2^{49}}\left| z \right| $
Now using $ z = a + ib $ then, $ \left| z \right| = \sqrt {{a^2} + {b^2}} $ property of complex number, we get
 $ {\left( {\sqrt {\sqrt {{{\left( 5 \right)}^2}} + \sqrt {{{\left( 3 \right)}^2}} } } \right)^{33}} = {2^{49}}\left| z \right| $
Now after solving inside bracket, we get
 $ {\left( {\sqrt 8 } \right)^{33}} = {2^{49}}\left| z \right| $
 $ {\left( 8 \right)^{\dfrac{{33}}{2}}} = {2^{49}}\left| z \right| $
 $ {({2^3})^{\dfrac{{33}}{2}}} = {2^{49}}\left| z \right| $
 $ {\left( 2 \right)^{\dfrac{{99}}{2}}} = {2^{49}}\left| z \right| $
 $ \left| z \right| = {2^{\left( {\dfrac{{99}}{2} - 49} \right)}} $
 $ \left| z \right| = {2^{\left( {\dfrac{{99 - 98}}{2}} \right)}} $
 $ \left| z \right| = {2^{\dfrac{1}{2}}} $
 $ \left| z \right| = \sqrt 2 $
The value of z modulus is $ \sqrt 2 $ . So, option (B) is the correct answer.
So, the correct answer is “Option B”.

Note: Whenever we face such types of problems we use some important points. First we find real and imaginary parts of a complex number, then apply the formula of modulus of a complex number, then after solving we can get the required answer. In this question we are required to find the modulus of z and in the question we are provided with an equation in which z is present. So, we took modulus on both sides to find the answer , the reason we took modulus is that, in the equation z was without modulus and we have to find the value of z modulus which is only possible with the way if we take modulus on both sides.