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UPSC NDA Math Model Paper 6

Question : 93 of 120

Marks: +1, -0

i^i

Solution:

Let

Z = i^i

As we know that, any complex number z = x + iy with modulus r and argument θ can also be expressed as

z = r \cdot e^{i\cdot\theta}

where

e^{i\theta} = \cos \theta + i \sin \theta.

∵

e^{i\frac{\pi}{2}} = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)

\Rightarrow e^{i\frac{\pi}{2}} = i

For the z = i, we have r = 1 and θ = π/2

Similarly, the given complex number Z can be expressed in the euler form as

i^i = \left(e^{i\frac{\pi}{2}}\right)^i = e^{i^2\frac{\pi}{2}} = e^{-\pi/2}

\Rightarrow Z = e^{-\pi/2}

∴ Imaginary part of

Z = i^i

is 0.

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