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sin[log(i^z)] with Least Modulus on |z−2i|=1 | JEE

JEE Maths question with a full step-by-step solution.

Question

The value of

\sin\!\left[\log_e\!\left\{\left(\cos\dfrac{\pi}{2} + i\sin\dfrac{\pi}{2}\right)^{\!z}\right\}\right]

, where

z

satisfies

|z - 2i| = 1

and has least modulus.

1

0

-1

correct

\dfrac{1}{2}

Solution

Step 1: Simplify the base. By Euler's formula,

\cos\dfrac{\pi}{2} + i\sin\dfrac{\pi}{2} = e^{i\pi/2} = i

. Step 2: Raise to the power

z

and take the logarithm:

\left(e^{i\pi/2}\right)^{z} = e^{i z\pi/2}, \qquad \log_e\!\left(e^{i z\pi/2}\right) = \frac{i z\pi}{2}

So the expression equals

\sin\!\left(\dfrac{i z\pi}{2}\right)

. Step 3: Find the value of

z

. The condition

|z - 2i| = 1

is a circle in the Argand plane centred at

2i

, i.e.,

(0, 2)

, with radius

1

. The point of least modulus (closest to the origin) lies on the line joining the origin to the centre, at distance

2 - 1 = 1

from the origin. That point is

z = i

. Step 4: Substitute

z = i

\sin\!\left(\frac{i\cdot i\cdot\pi}{2}\right) = \sin\!\left(\frac{i^2\pi}{2}\right) = \sin\!\left(-\frac{\pi}{2}\right) = -1

Correct answer: (3)

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