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Question: If \( {\left( {\sqrt 5 + \sqrt {3i} } \right)^{33}} = {2^{49}}z \) , then modulus of the complex num...

If (5+3i)33=249z{\left( {\sqrt 5 + \sqrt {3i} } \right)^{33}} = {2^{49}}z , then modulus of the complex number is
A.1
B. 2\sqrt 2
C. 222\sqrt 2
D.4

Explanation

Solution

Hint : The complex number in the question is z so we have to find modulus of z which is ∣z∣\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: (5+3i)33=249z{\left( {\sqrt 5 + \sqrt {3i} } \right)^{33}} = {2^{49}}z
Now, we have to find modulus of z which is ∣z∣\left| z \right| , so
By taking modulus on both sides we get,
∣(5+3i)33∣=∣249z∣\left| {{{\left( {\sqrt 5 + \sqrt {3i} } \right)}^{33}}} \right| = \left| {{2^{49}}z} \right|
By using the property of complex number ∣zn∣=∣z∣n\left| {{z^n}} \right| = {\left| z \right|^n} , we get
∣(5+3i)∣33=249∣z∣{\left| {\left( {\sqrt 5 + \sqrt {3i} } \right)} \right|^{33}} = {2^{49}}\left| z \right|
Now using z=a+ibz = a + ib then, ∣z∣=a2+b2\left| z \right| = \sqrt {{a^2} + {b^2}} property of complex number, we get
((5)2+(3)2)33=249∣z∣{\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
(8)33=249∣z∣{\left( {\sqrt 8 } \right)^{33}} = {2^{49}}\left| z \right|
(8)332=249∣z∣{\left( 8 \right)^{\dfrac{{33}}{2}}} = {2^{49}}\left| z \right|
(23)332=249∣z∣{({2^3})^{\dfrac{{33}}{2}}} = {2^{49}}\left| z \right|
(2)992=249∣z∣{\left( 2 \right)^{\dfrac{{99}}{2}}} = {2^{49}}\left| z \right|
∣z∣=2(992−49)\left| z \right| = {2^{\left( {\dfrac{{99}}{2} - 49} \right)}}
∣z∣=2(99−982)\left| z \right| = {2^{\left( {\dfrac{{99 - 98}}{2}} \right)}}
∣z∣=212\left| z \right| = {2^{\dfrac{1}{2}}}
∣z∣=2\left| z \right| = \sqrt 2
The value of z modulus is 2\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.