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

Last updated date: 04th Mar 2024
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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.

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.