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# GATE-2012 ECE Q25 (math)

by on October 9, 2012

Question 25 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

## Q25. If $\Large{x=\sqrt{-1}}$, then the value of $\Large{x^x}$ is,

(a) $e^{-\pi/2}$

(b) $e^{\pi/2}$

(c) $x$

(d) 1

## Solution

Going by the definition of imaginary unit$j^2 =-1$ and

From Euler’s formula$e^{jy}=\cos(y) + j\sin(y)$.

Putting $y=\pi/2$,

$e^{j\pi/2}=\cos(\pi/2) + j\sin(\pi/2) = 0 + 1j = j$.

Substituting,

$\Large{x^x=j^j=e^{$$j\pi/2$$^j}=e^{-\pi/2}}$.

## References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

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