Find all complex numbers z such that \(z^2 = i\)

I can't write sqrt(i), so I don't know what the answer is.

Guest Jul 5, 2020

#1**+1 **

If you are wondering how to write the sqrt of i : (this is NOT the answer to the question though see alan's answer below))

i = sqrt (-1) = -1^{1/2}

the square root of this would be

-1^{1/4 }or \(\sqrt[1/4]{-1}\)

ElectricPavlov Jul 5, 2020

#2**+2 **

Find all complex numbers z z^2=i

1) \(i=\sqrt {-1}\)

2) Euler:

\(e^{i\pi}=-1\\ \sqrt{e^{i\pi}}=\sqrt{-1 } \\ e^{i \frac{\pi}{ 2}}=i\)

\(\sqrt{ e^{ i \frac{\pi}{2} } } =\pm \sqrt{i} \\ e^{ i\frac{\pi}{4} } =\pm \sqrt{i} \\ \sqrt{i} = \pm e^{i\frac{\pi}{4} } \)

\(z^2=i \\ z= \sqrt{i}\)

\(z_1= + e^{i\frac{\pi}{4}} \\= \cos(45^\circ) + i \sin(45^\circ) \\ = \frac{\sqrt{2}} {2}+i \frac{\sqrt2} {2} \)

\(z_2= -e^{i\frac{\pi}{4}} \\=- \frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\)

heureka Jul 5, 2020