joealanis98 joealanis98
  • 11-05-2017
  • Mathematics
contestada

For z = 2cis Pi/6 , find z5 in rectangular form.

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LammettHash
LammettHash LammettHash
  • 11-05-2017
[tex]z=2\mathrm{cis}\dfrac\pi6=2\left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)[/tex]

By DeMoivre's theorem, you have

[tex]z^5=2^5\mathrm{cis}\dfrac{5\pi}6=32\left(\cos\dfrac{5\pi}6+i\sin\dfrac{5\pi}6\right)[/tex]
[tex]\implies z^5=32\left(-\dfrac{\sqrt3}2+\dfrac12i\right)=-16\sqrt3+16i[/tex]
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