Given z=i−12. Evaluate z1/2 and show the roots.
I got z1/2=121/4(cos3π8+isin3π8)
First of all is my z1/2 correct? And secondly how to show the roots??
I would be glad and appreciated if someone could help me out.
Answer
z=−12+i2=R(cosy+isiny)(say) where R>0
Rcosy=−12,Rsiny=12⟹R2=(−12)2+(12)2=12
So, tany=RsinyRcosy=−1⟹y=2nπ+3π4 as cosy<0 and siny>0 ,so y lies in the 2nd quadrant.
So, z=1√2(cos3π4+isin3π4)
So, the general value of z12 is (12)14(cos(nπ+3π8)+isin(nπ+3π8)) where n=0,1
using de Moivre's formula.
n=0⟹z12=(12)14(cos(3π8)+isin(3π8))
n=1⟹z12=(12)14(cos(π+3π8)+isin(π+3π8))=−(12)14(cos(3π8)+isin(3π8))
One may look into this, for the values of n.
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