If z = 2cis120°, find z2 in polar form?

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Answer 1 · 2018-05-17T10:52:06

We know #cis(x)=cos(x)+isin(x)#

So

#z=2cis(120^@)#

#=>z=2(cos(120^@)+isin(120^@))#

#=>z=2(cos(180^@-60^@)+isin(180^@-60^@))#

#=>z=2(-cos(60^@)+isin(60^@))#

#=>z=2(-1/2+isqrt3/2)#

#=>z=-1+isqrt3#

#=>z^2=(-1+isqrt3)^2#

#=>z^2=1-3-i2sqrt3#

#=>z^2=-2-i2sqrt3#

#=>z^2=4(-2/4-i(2sqrt3)/2)#

#=>z^2=4(-1/2-isqrt3/2)#

#=>z^2=4(cos240^@+i sin240^@)#

If z = 2cis120°​, find z2 in polar form?