If z = 32 + i2, then (z^201 - i)^8 is equal to — Solution | JEE Mathematics

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If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$ , then $(z^{201} - i)^8$ is equal to

Answer

Correct answer:B

Standard Method

Given: $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$

Find: The value of $(z^{201} - i)^8$

Write $z$ in trigonometric form:

z = \cos 30^\circ + i \sin 30^\circ = e^{i\pi/6}

Now,

z^{201} = e^{i\,201\pi/6} = e^{i\,67\pi/2}

Reduce the angle:

\frac{67\pi}{2} = 32\pi + \frac{3\pi}{2}

So,

z^{201} = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right) = -i

Therefore,

(z^{201} - i)^8 = (-i - i)^8 = (-2i)^8

Now use powers of $i$ :

(-2i)^8 = (-2)^8 i^8 = 256 \times 1 = 256

Therefore, the correct option is B and the value is $256$ .

Writing $z$ in the wrong trigonometric form. Here $\frac{\sqrt{3}}{2} = \cos 30^\circ$ and $\frac{1}{2} = \sin 30^\circ$ , so $z = \cos 30^\circ + i\sin 30^\circ$ . Do not interchange sine and cosine; first identify the correct angle.
Not reducing the exponent angle modulo $2\pi$ . After computing $z^{201} = e^{i67\pi/2}$ , the angle must be reduced to an equivalent principal angle. Use $\frac{67\pi}{2} = 32\pi + \frac{3\pi}{2}$ before evaluating sine and cosine.
Handling powers of $i$ incorrectly. Since $i^4 = 1$ , we get $i^8 = (i^4)^2 = 1$ . Do not stop at $(-2i)^8$ without simplifying the power of $i$ .

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