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GATE-2012 ECE Q25 (math)

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

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

(a) $e^{-\pi/2}$

(b) $e^{\pi/2}$

(d) 1

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,

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

[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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