Wednesday, 9 March 2016

Solving roots of complex number



Given z=i12. 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=12R2=(12)2+(12)2=12



So, tany=RsinyRcosy=1y=2nπ+3π4 as cosy<0 and siny>0 ,so y lies in the 2nd quadrant.



So, z=12(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=0z12=(12)14(cos(3π8)+isin(3π8))



n=1z12=(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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