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

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

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Answer 1 · 2016-12-17T14:22:03

$z = \frac{\sqrt{2}}{2} (1 + i)$ $z=sqrt2/2(1+i)$ , $z^{2} = i$ $z^2=i$ , $z^{3} = \frac{\sqrt{2}}{2} (- 1 + i)$ $z^3=sqrt2/2(-1+i)$ and $z^{4} = - 1$ $z^4=-1$

Explanation:

$z = \frac{\sqrt{2}}{2} (1 + i)$ $z=sqrt2/2(1+i)$

= $(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2})$ $(sqrt2/2+isqrt2/2)$ and in trigonometric form

= $(cos (\frac{π}{4}) + i sin (\frac{π}{4}))$ $(cos(pi/4)+isin(pi/4))$ or in exponential

= $e^{i \frac{π}{4}}$ $e^(ipi/4)$

Hence $z^{2} = e^{i \frac{2 π}{4}} = e^{i \frac{π}{2}} = (cos (\frac{π}{2}) + i sin (\frac{π}{2})) = i$ $z^2=e^(i(2pi)/4)=e^(ipi/2)=(cos(pi/2)+isin(pi/2))=i$

and $z^{3} = e^{i \frac{3 π}{4}} = (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4})) = - \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$ $z^3=e^(i(3pi)/4)=(cos((3pi)/4)+isin((3pi)/4))=-sqrt2/2+isqrt2/2$

= $\frac{\sqrt{2}}{2} (- 1 + i)$ $sqrt2/2(-1+i)$

and

$z^{4} = e^{i \frac{4 π}{4}} = e^{i π} = (cos π + i sin π) = - 1$ $z^4=e^(i(4pi)/4)=e^(ipi)=(cospi+isinpi)=-1$

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

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