1 answer
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first find the number 3-i in polar form
r = β(3^2 + 1^2) = β10
tanΓ = (-1/3) , where Γ is in IV
Γ = 341.565Β°
so 3-i = β10(cos 341.565 + isin341.565)
then
(3-i)^(1/2) = [ (β10)(cos 341.565 + isin341.565) ]^(1/2)
= β10^(1/2)(cos 170.7825 + isin 170.7825) by DeMoirve's Theorem
finally (5/2)β(3-1)
= 5(10)^(1/4)/2)[cos 170.8Β° + isin 170.8Β°)
or
5(10)^(1/4) cis 170.8Β°