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Square root involving complex number

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  • #1
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 [tex] C = 1/\sqrt{i*x} [/tex]. find the two roots of C. The solution for one of the roots is given as [tex] C = \sqrt{1/2x} - i\sqrt{1/2x} [/tex] 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:

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

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 [tex] \pi/2 [/tex] so I get:

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

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

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

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

[tex] C = \frac{\sqrt{x}}{2x} - i \frac{1}{2x} [/tex]

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?
 

Answers and Replies

  • #2
gabbagabbahey
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1. The complex number (C) is [tex] C = 1/\sqrt{i*x} [/tex]. 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 [itex]C^2=\frac{1}{ix}[/itex]? The equation you've given is linear in [itex]C[/itex] and hence has only one solution. The equation that I gave is quadratic and hence has two solutions, [itex]C=\pm\frac{1}{\sqrt{ix}}[/itex]

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

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

[tex] C = \frac{\sqrt{x}}{2x} - i \frac{1}{2x} [/tex]

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?
[tex]\cos\left(\frac{\pi}{4}\right)=\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\neq \frac{1}{2}[/tex]
 
  • #3
But there is only one solution to this equation...do you mean that you are trying to find the roots of the equation [itex]C^2=\frac{1}{ix}[/itex]? The equation you've given is linear in [itex]C[/itex] and hence has only one solution. The equation that I gave is quadratic and hence has two solutions, [itex]C=\pm\frac{1}{\sqrt{ix}}[/itex]
No, the problem definitely says [tex]C=\frac{1}{\sqrt{ix}}[/tex] 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...


[tex]\cos\left(\frac{\pi}{4}\right)=\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\neq \frac{1}{2}[/tex]
I know, the [tex] \frac{1}{2} [/tex] is from [tex] cos^2(\frac{\pi}{4}) [/tex] from multiplying with the complex conjugate.

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

can be expressed as:

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

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:

[tex] 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 [/tex]

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
gabbagabbahey
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No, the problem definitely says [tex]C=\frac{1}{\sqrt{ix}}[/tex] 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 [tex] \frac{1}{2} [/tex] is from [tex] cos^2(\frac{\pi}{4}) [/tex] from multiplying with the complex conjugate.

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

[tex]\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}}[/tex]

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

can be expressed as:

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

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:

[tex] 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 [/tex]
[/QUOTE]

If they are finding [itex]c^{1/2}[/itex] instead of [itex]c[/itex], 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 [itex]c=\frac{1}{ix}[/itex] not [itex]c=\frac{1}{\sqrt{ix}}[/itex]. And by "finding the two roots of [itex]c[/itex]", they mean to find the two numbers [itex]c^{1/2}[/itex] such that [itex]c=\frac{1}{ix}[/itex].
 
Last edited:

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