Square root involving complex number

[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:

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

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?

 

[gabbagabbahey]

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}

 

[hypothesis]

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]

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

[/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}.