Question

Find the square roots of 36 + 36sqrt3i

Answer 1

`3√(6)+3√(2)i` and
`-3√(6)-3√(2)i`

Explanation:

For a complex number
`z=a+bi` in rectangular form, polar form is
`z=r[cos(\theta)+isin(\theta)]` where
`r=√(a^2+b^2)` and
`\theta=tan^(-1)((b)/(a))`:

`r=√(a^2+b^2)\\\\r=\sqrt{(36)^2+(36√(3))^2}\\\\r=√(1296+3888)\\\\r=√(5184)\\\\r=72`

`\theta=tan^(-1)((b)/(a))\\ \\\theta=tan^(-1)((36√(3))/(36))\\ \\\theta=tan^(-1)(√(3))\\\\\theta=(\pi)/(3)`

Thus, our complex number in polar form is
`z=72[cos((\pi)/(3))+isin((\pi)/(3))]`. This will help us determine the square roots of the complex number using the formula
`\sqrt[n]{r}\biggr[cis\bigr({(\theta+2\pi k)/(n))\biggr]` for
`k=0,1,2,\: ... \:,n-1`. Note that
`cis(\theta)=cos(\theta)+isin(\theta)`:

1st square root for k = 2-1 = 1

`\sqrt[n]{r}\biggr[cis\bigr({(\theta+2\pi k)/(n))\biggr]`

`\sqrt[2]{72}\biggr[cis\bigr({((\pi)/(3)+2\pi(1))/(2))\biggr]`

`6√(2)\biggr[cis\bigr({((\pi)/(3)+2\pi)/(2))\biggr]`

`6√(2)\biggr[cis\bigr((\pi)/(6)+\pi)\biggr]`

`6√(2)\biggr[cis\bigr((7\pi)/(6))\biggr]`

`6√(2)\biggr[cos((7\pi)/(6))+isin((7\pi)/(6))\biggr]\\ \\6√(2)(-(√(3))/(2)-(1)/(2)i)\\\\-3√(6)-3√(2)i`

2nd square root for k = 1-1 = 0

`\sqrt[6]{72}\biggr[cis\bigr({((\pi)/(3)+2\pi(0))/(2)\bigr)\biggr]`

`6√(2)\biggr[cis((\pi)/(6) )\biggr]`

`6√(2)[cos((\pi)/(6))+isin((\pi)/(6))]\\ \\6√(2)((√(3))/(2)+(1)/(2))\\\\3√(6)+3√(2)i`

Therefore, the square roots of
`36+36√(3)i` are
`3√(6)+3√(2)i` and
`-3√(6)-3√(2)i`.