How do you find $z, z^2, z^3, z^4$ given $z=sqrt2/2(1+i)$ ?

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Answer 1 · 2016-12-17T14:22:03

$z=sqrt2/2(1+i)$ , $z^2=i$ , $z^3=sqrt2/2(-1+i)$ and $z^4=-1$

$z=sqrt2/2(1+i)$

= $(sqrt2/2+isqrt2/2)$ and in trigonometric form

= $(cos(pi/4)+isin(pi/4))$ or in exponential

= $e^(ipi/4)$

Hence $z^2=e^(i(2pi)/4)=e^(ipi/2)=(cos(pi/2)+isin(pi/2))=i$

and $z^3=e^(i(3pi)/4)=(cos((3pi)/4)+isin((3pi)/4))=-sqrt2/2+isqrt2/2$

= $sqrt2/2(-1+i)$

and

$z^4=e^(i(4pi)/4)=e^(ipi)=(cospi+isinpi)=-1$

How do you find z, z^2, z^3, z^4z, z^2, z^3, z^4 given z=sqrt2/2(1+i)z=sqrt2/2(1+i)?