# Square root involving complex number

1. Mar 15, 2010

### hypothesis

I've struggled for days reading about square roots of complex numbers and I get most of the problems but not this one. I really want to understand what is going on in this problem, hope someone can help!

1. The complex number (C) is $$C = 1/\sqrt{i*x}$$. find the two roots of C. The solution for one of the roots is given as $$C = \sqrt{1/2x} - i\sqrt{1/2x}$$ can someone show me how to get to that solution?

2. Equations: the relationships between polar and rectangular coordinates for complex variables

3. I try taking the square root of i*x by using the relation:

$$\sqrt{a+ib} = \sqrt{r}\left(cos(\theta/2)+i*sin(\theta/2))$$

Then substitute with my variables (a+ib) = (0+ix) and because the real part is zero and the imaginary part is positive the angle is $$\pi/2$$ so I get:

$$\sqrt{0+ix} = \sqrt{\sqrt{0^2+x^2}}(cos(\pi/4)+i*sin(\pi/4))$$
=
$$\sqrt{ix} = \sqrt{x}(cos(\pi/4)+i*sin(\pi/4)$$

Substitute it back into the first equation ($$C = 1/\sqrt{i*x}$$) and get:

$$C = \frac{1}{\sqrt{x}(cos(\pi/4)+i*sin(\pi/4)}$$

now I can multiply nominator and denominator with the conjugate of the denominator and come up with a solution:

$$C = \frac{\sqrt{x}}{2x} - i \frac{1}{2x}$$

but this is not the solution that I have been given plus I was supposed to find two solutions so I'm clearly doing something wrong. But what?

2. Mar 15, 2010

### gabbagabbahey

But there is only one solution to this equation...do you mean that you are trying to find the roots of the equation $C^2=\frac{1}{ix}$? The equation you've given is linear in $C$ and hence has only one solution. The equation that I gave is quadratic and hence has two solutions, $C=\pm\frac{1}{\sqrt{ix}}$

$$\cos\left(\frac{\pi}{4}\right)=\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\neq \frac{1}{2}$$

3. Mar 15, 2010

### hypothesis

No, the problem definitely says $$C=\frac{1}{\sqrt{ix}}$$ and that there are two roots. Maybe a typo in the book then, I don't know, but if that is the case I have wasted a lot of time...

I know, the $$\frac{1}{2}$$ is from $$cos^2(\frac{\pi}{4})$$ from multiplying with the complex conjugate.

In the text they continue by stating that the polar form of
$$\frac{1}{ix} = \frac{-i}{x}$$

can be expressed as:

$$c = \frac{1}{x}\left(cos\left(\frac{3\pi}{2}\right) + i*sin\left(\frac{3\pi}{2}\right)\right)$$

I understand that this is true but I don't understand how they just disregard the square root that was in the denominator. They then go on to find c^(1/2) like this:

$$c^\frac{1}{2} = \sqrt{\frac{1}{x}}\left(cos\left(\frac{3\pi}{4} + k*\pi\right) + i*sin\left(\frac{3\pi}{4} + k*\pi\right)\right) ; k = 0,1$$

and I guess this equation has two solutions but I still don't get how they get there.. does this make sense to you, or anyone?

thanks for the fast reply gabbagabbahey.

4. Mar 15, 2010

### gabbagabbahey

It's either a typo, or the book you are using has a multivalued definition of the square root sign (which I've never seen before).

Okay, but you will have a $\sin$ and a $\cos$ in the numerator then and so there should be a $\sqrt{2}$ in there somewhere.

$$\frac{1}{\cos(\pi/4)+i\sin(\pi/4)}= \frac{1}{\cos(\pi/4)+i\sin(\pi/4)}\left(\frac{\cos(\pi/4)-i\sin(\pi/4)}{\cos(\pi/4)-i\sin(\pi/4)}\right)=\frac{\cos(\pi/4)-i\sin(\pi/4)}{\cos^2(\pi/4)+\sin^2(\pi/4)}=\frac{\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}}{\frac{1}{2}+\frac{1}{2}}=\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}$$

[/QUOTE]

If they are finding $c^{1/2}$ instead of $c$, then the problem statement must be a typo (either that, or the solutions they give are very incorrect). If I were you, I would assume that the question was meant to $c=\frac{1}{ix}$ not $c=\frac{1}{\sqrt{ix}}$. And by "finding the two roots of $c$", they mean to find the two numbers $c^{1/2}$ such that $c=\frac{1}{ix}$.

Last edited: Mar 15, 2010