# Square root involving complex number

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?

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gabbagabbahey
Homework Helper
Gold Member
1. The complex number (C) is $$C = 1/\sqrt{i*x}$$. find the two roots of C.

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, $C=\pm\frac{1}{\sqrt{ix}}$

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

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

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, $C=\pm\frac{1}{\sqrt{ix}}$
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...

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

gabbagabbahey
Homework Helper
Gold Member
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...
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).

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

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.

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

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:

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

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