complex numbers

**joey1** · June 11th 2009, 11:56 PM

what are the roots of square root of 1-1.73j giving results in the rectangular and in exponential form

**Twig** · June 12th 2009, 12:49 AM

hi

I am not sure exactly what you mean, but if $z \in \mathbb{C} \; z=a+bi$ , then $|z| = \sqrt{a^{2}+b^{2}}$

And the argument of the complex number would be:

$2\pi - arctan(1.73)$ , since you are in the fourth quadrant. Do you understand why here?

A complex number can be expressed: $|z|(cos(\theta)+i\cdot sin(\theta)$ where $\theta$ is the argument of z, or equivalently $e^{ln(|z|)}(cos(\theta)+i \cdot sin(\theta)$

**HallsofIvy** · June 12th 2009, 02:38 AM

Is that really 1.73 or is it $\sqrt{3}$ ? I am going to assume you mean $\sqrt{3}$ .

Once you have $z= 1- i\sqrt{3}$ in polar form, $z= r(cos(\theta)+ i sin(\theta))$ , which is what Twig gave you, its square roots are $r^{1/2}(cos(\theta/2)+ i sin(\theta/2)$ and $r^{1/2}(cos(\theta/2+ \pi)+ i sin(\theta/2+ \pi)$ .

Here, $r^{1/2}$ is the "principle" root of the real number r: for this problem $r= \sqrt{1+ (\sqrt{3})^2}= 2$ and $\theta= tan^{-1}(-\sqrt{3})= -\pi/3$ . (Twig's value was $2\pi- \pi/3= 5\pi/3$ but I prefer $-\pi/3$ . They give the same z.)

So the square roots of $1- i\sqrt{3}$ are $\sqrt{2}(cos(-\pi/6)+ i sin(-\pi/6))= \sqrt{2}(\frac{1}{2}- \frac{\sqrt{3}}{2}i)$ and $\sqrt{2}(cos(5\pi/6)+ i sin(5\pi/6))= \sqrt{2}(-\frac{1}{2}+ \frac{\sqrt{3}}{2}i)$ .

(Sorry, but I just can't bring myself to write "j" instead of "i"! Oh, those engineers and their jmaginary numbers!)

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