what are the roots of square root of 1-1.73j giving results in the rectangular and in exponential form
hi
I am not sure exactly what you mean, but if $\displaystyle z \in \mathbb{C} \; z=a+bi $ , then $\displaystyle |z| = \sqrt{a^{2}+b^{2}} $
And the argument of the complex number would be:
$\displaystyle 2\pi - arctan(1.73) $ , since you are in the fourth quadrant. Do you understand why here?
A complex number can be expressed: $\displaystyle |z|(cos(\theta)+i\cdot sin(\theta) $ where $\displaystyle \theta $ is the argument of z, or equivalently $\displaystyle e^{ln(|z|)}(cos(\theta)+i \cdot sin(\theta) $
Is that really 1.73 or is it $\displaystyle \sqrt{3}$? I am going to assume you mean $\displaystyle \sqrt{3}$.
Once you have $\displaystyle z= 1- i\sqrt{3}$ in polar form, $\displaystyle z= r(cos(\theta)+ i sin(\theta))$, which is what Twig gave you, its square roots are $\displaystyle r^{1/2}(cos(\theta/2)+ i sin(\theta/2)$ and $\displaystyle r^{1/2}(cos(\theta/2+ \pi)+ i sin(\theta/2+ \pi)$.
Here, $\displaystyle r^{1/2}$ is the "principle" root of the real number r: for this problem $\displaystyle r= \sqrt{1+ (\sqrt{3})^2}= 2$ and $\displaystyle \theta= tan^{-1}(-\sqrt{3})= -\pi/3$. (Twig's value was $\displaystyle 2\pi- \pi/3= 5\pi/3$ but I prefer $\displaystyle -\pi/3$. They give the same z.)
So the square roots of $\displaystyle 1- i\sqrt{3}$ are $\displaystyle \sqrt{2}(cos(-\pi/6)+ i sin(-\pi/6))= \sqrt{2}(\frac{1}{2}- \frac{\sqrt{3}}{2}i)$ and $\displaystyle \sqrt{2}(cos(5\pi/6)+ i sin(5\pi/6))= \sqrt{2}(-\frac{1}{2}+ \frac{\sqrt{3}}{2}i)$.
(Sorry, but I just can't bring myself to write "j" instead of "i"! Oh, those engineers and their jmaginary numbers!)