Question

what is the sq rt of i (i = sq rt of -1)

Answer 1

(a + bi)^2 = i

Answer 2

\[\sqrt{i} = (i)^{1/2} \]
\[z = r(cos\theta + isin\theta)\]
\[i => z = 1(cos(\frac{pi}{2})+ isin\frac{pi}{2})\]

using demoivre's
\[\sqrt{i} => (cos(pi/2) + isin(pi/2))^{1/2}\]
\[z=(cos(\frac{pi}{4})+isin(\frac{pi}{4})\]