+0  
 
0
118
2
avatar

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
avatar+28021 
+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) =   -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
avatar+25597 
+2

Find all complex numbers z z^2=i

 

1)  \(i=\sqrt {-1}\)

 

2) Euler:

 

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

 

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

 

 

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

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

 

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

 

 

laugh

   

 
 Jul 5, 2020

44 Online Users

avatar