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Find all complex numbers z that satisfy z^2 = i/2.

 Feb 3, 2022
 #1
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Put the number into polar form \(\displaystyle (1/2,\pi/2), \)  (abbreviated).

We're finding a root so the number has to be put into its multivalued form, \(\displaystyle (1/2,\pi/2+2k\pi), k=0,1,2,...\)

\(\displaystyle z=(1/2,\pi/2+2k\pi)^{1/2},k=0,1,\\=(1/\sqrt{2},\pi/4+k\pi),k=0,1,\\ =(k=0),(1/\sqrt{2},\pi/4)=(1/2+i/2),\\ \text{or }(k=1),(1/\sqrt{2},5\pi/4)=(-1/2-i/2).\)

 Feb 4, 2022

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