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# complex numbers

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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.

Jul 5, 2020

#1
+28021
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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) =   -11/2

the square root of this would be

-11/4         or      $$\sqrt[1/4]{-1}$$

Jul 5, 2020
edited by ElectricPavlov  Jul 5, 2020
#2
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Find all complex numbers z z^2=i

1)  $$i=\sqrt {-1}$$

2) Euler:

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

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

$$z^2=i \\ z= \sqrt{i}$$

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

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

Jul 5, 2020