Imaginary sum

Algebra Level 3

$\large \sqrt{i} + \sqrt{-i} = \, ?$

i

-i

0

\sqrt2

1 solution

Abhay Tiwari
Apr 27, 2016

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

$\sqrt{i}=\sqrt{e^{\frac{\pi}{2}}}$

$\sqrt{i}=e^{\frac{\pi}{4}}=\cos(\frac{\pi}{4})+i \sin(\frac{\pi}{4})$

similarly, $\sqrt{i}=e^{-\frac{\pi}{4}}= \cos(-\frac{\pi}{4})+i \sin(-\frac{\pi}{4})$

Then, $\sqrt{i}+\sqrt{-i}= \cos(\frac{\pi}{4})+i \sin(\frac{\pi}{4})+ \cos(-\frac{\pi}{4})+i \sin(-\frac{\pi}{4})=\sqrt{2}$

Perfect solution!

Ashish Menon - 4 years, 11 months ago

I don't believe that i = e^(pi/2), since i is "imaginary", e^(pi/2) is real. Actually, e^(i (pi/2) = cos(pi/2) + i sin(pi/2) = 0 + i = i, and e^(-i pi/2) = cos(pi/2) - i sin(pi/2) = -i. Therefore sqrt(i) + sqrt(-i) = e^(i pi/4) + e^(-i pi/4) = cos(pi/4) + i sin(pi/4) + cos(pi/4) - i sin(pi/4) = 2 cos(pi/4) = 2 sqrt(2)/2 = sqrt(2).

Edwin Gray - 2 years, 2 months ago

Imaginary sum

1 solution

1 pending report

Vote up reports you agree with