Question

What is (1/2 + isqrt3/2)^5?

Answer 1

`((1)/(2)+(√(3))/(2)i)^5=(1)/(2)-(√(3))/(2)i`

Explanation:

Convert 1/2 + i√3/2 to rectangular form

`\displaystyle z=a+bi=(1)/(2)+(√(3))/(2)i\\\\r=√(a^2+b^2)=\sqrt{\biggr((1)/(2)\biggr)^2+\biggr((√(3))/(2)\biggr)^2}=\sqrt{(1)/(4)+(3)/(4)}=√(1)=1\\\\\theta=\tan^(-1)\biggr((b)/(a)\biggr)=\tan^(-1)\biggr(((√(3))/(2))/((1)/(2))\biggr)=\tan^(-1)(√(3))=(\pi)/(3)\\\\z=\cos(\pi)/(3)+i\sin(\pi)/(3)`

Use DeMoivre's Theorem

`\displaystyle z^n=r^n(\cos(n\theta)+i\sin(n\theta))\\\\z^5=1^5\biggr(\cos\biggr((5\pi)/(3)\biggr)+i\sin\biggr((5\pi)/(3)\biggr)\biggr)\\\\z^5=(1)/(2)-(√(3))/(2)i`