If z=≤ft({{3}}{2}+i/2)^5+≤ft({{3}}{2}-i/2)^5, then

Correct Answer is : Im⁡(z)=0{Im}(z)=0Im(z)=0

Correct Answer is : Im⁡(z)=0{Im}(z)=0Im(z)=0

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COMEDK 2025 Evening Solved Paper

Question : 142 of 180

Marks: +1, -0

z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5

, then

Solution:

Concept:

Express each complex number in Euler’s form to simplify the sum of powers.

Explanation:

Let

a = \frac{\sqrt{3}}{2} + \frac{i}{2}

Since

\cos(\pi/6) = \frac{\sqrt{3}}{2}

and

\sin(\pi/6) = \frac{1}{2}

, we have

a = e^{i\pi/6}

Similarly,

\bar{a} = \frac{\sqrt{3}}{2} - \frac{i}{2} = e^{-i\pi/6}

Then

z = a^5 + \bar{a}^5 = e^{i5\pi/6} + e^{-i5\pi/6}

Using Euler’s formula,

e^{i\theta} + e^{-i\theta} = 2\cos\theta

z = 2\cos(5\pi/6)

Now

\cos(5\pi/6) = -\frac{\sqrt{3}}{2}

, therefore

z = 2\left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3}

Since

-\sqrt{3}

is purely real, its imaginary part is zero.

Answer:

\operatorname{Im}(z) = 0

, which corresponds to option B.

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