Square root of complex number in polar or rectangular form

Question

I am trying to find how to simplify:

$$\sqrt{\frac{A+jb}{C+jd}}$$

My calculator errors out, giving a math error, and I don't know how else to solve this.

Answer 1

Thanks Soham, and others for your efforts, but here is a much straight-forward just pluck in the values solution to my problem:

Answer 2

Hint

It suffices to know how to find $\alpha+i\beta$ such that $(\alpha+i\beta)^2=a+ib$.

Expand the last equality then solve for $\alpha$ and $\beta$ the equations: $$2\alpha\beta=b\quad\text{and}\quad \alpha^2-\beta^2=a$$

Answer 3

I understand you're trying to convert to polar.

Multiply top and bottom by $\overline{C-id}:$ $$\sqrt{\frac{(A+ib)(C-id)}{(C+id)(C-id)}} = \sqrt{\frac{(AC+bd) + i(bC - Ad)}{C^2 + D^2}}$$

Work from that! Let $z$ = the complex number $(AC+bd) + i(bC - Ad)$.

Then, convert $z$ to polar form:

Firstly, $\sqrt{(AC+bd)^2 + (bC - Ad)^2}$ can be simplified to $\sqrt{(A^2 + b^2)(C^2+d^2)}$.

The argument of $z$ is $$\tan^{-1}\bigg(\frac{bC-Ad}{AC+bd}\bigg)$$ Hence $z$ can be expressed as $$\bigg[\sqrt{(A^2 + b^2)(C^2+d^2)}\bigg]\exp\bigg({i\tan^{-1}\bigg(\frac{bC-Ad}{AC+bd}\bigg)\bigg)}$$

Therefore: $$\sqrt{\frac{(AC+bd) + i(bC - Ad)}{C^2 + d^2}}$$$$ = \sqrt{\frac{\bigg[\sqrt{(A^2 + b^2)(C^2+d^2)}\bigg]\exp\bigg({i\tan^{-1}\bigg(\frac{bC-Ad}{AC+bd}\bigg)\bigg)}}{C^2 + d^2}}$$$$ = \exp\bigg({\frac{1}{2}i\tan^{-1}\bigg(\frac{bC-Ad}{AC+bd}\bigg)\bigg)}\sqrt{\frac{\sqrt{(A^2 + b^2)(C^2+d^2)}}{C^2 + d^2}} $$$$ = \bigg(\frac{A^2 + b^2}{C^2+d^2}\bigg)^{1/4}\exp\bigg({\frac{1}{2}i\tan^{-1}\frac{bC-Ad}{AC+bd}\bigg)}$$

Fine enough?