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Math Help - nth root of a complex number

  1. #1
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    nth root of a complex number

    Ok so basically I'm supposed to understand complex numbers (so that we can use them for eigenvalues and eigenvectors), but then someone dropped the ball and we got almost no teaching on the subject and now pretty much everyone in my class who hasn't worked with complex numbers before (which includes me) are having trouble understanding.

    Ok, so I'm doing an example question and I just can't seem to get it to work:

    z^4 = -64 and I have to find z. Immediately my response was to find the fourth root of 64 ( 2\sqrt{2}) and multiply it by some form of i whose fourth power would be equal to negative 1. This turned out to be \sqrt{i} and I'm sure that there's various terms that could be made positive or negative (or some duplicate terms) yielding four roots. Great. I still don't understand it, that was essentially guess and check.

    So my teacher did show us a method for finding the nth root of a complex number (I'll cut out most of the algebraic steps as I'm sure most of the people who can help me will be well aware of this method):
    So if you have
    z^n = w, where w is a complex number, you convert w into polar form and then set
    z = te^{(i\phi)} so that you get
    (te^(i\phi))^n = re^{(i\theta)}
    After some algebraic manipulation, you get
    z = \sqrt[n]{r} e^{(i(\frac{\theta + m2\pi}{n}))} where m is an integer.

    So I tried that with z^4 = -64 = -64 + 0i. To convert, for the modulus I got r = \sqrt{(-64)^2 + (0i)^2} = 64 and for the argument I got \theta = arctan(\frac{0}{-64}) = 0. So...
    z = \sqrt[4]{64} e^{(i(\frac{0 + m2\pi}{4}))} <br />
= 2\sqrt{2} e^{(i(\frac{m\pi}{2}))} and when you change that back into algebraic form (the form the answer is required in) you get(for m=0) z = 2\sqrt{2}(\cos0 + i\sin0) = 2\sqrt{2} which is clearly not right. What am I doing wrong? The problem I see is that when I convert a pure real number into a complex number in polar form, the polar form is the same regardless of whether the number is positive or negative. Any help with solving problems of this type is greatly appreciated.
    Last edited by mr fantastic; November 6th 2010 at 01:47 PM.
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  2. #2
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     - 64 = 64\exp \left( {i\pi } \right)

    Let \sigma  = \sqrt[n]{{64}}\exp \left( {\frac{{i\pi }}{n}} \right)\;\& \,\xi  = \exp \left( {\frac{{2i\pi }}{n}} \right).

    The roots are \sigma\cdot \xi^k,~k=0,1,\cdots,n-1.
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  3. #3
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    How did you get pi as the argument? We were taught that if you have a complex number z, then
    z = x + yi = r \exp(i\theta) where r = \sqrt{x^2 + y^2} and \theta = \arctan \frac{y}{x}

    So in this case wouldn't \theta = \arctan \frac{0}{-64} = \arctan 0 = 0 ?
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  4. #4
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    For any negative real number, \pi is the argument.
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  5. #5
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    Interesting. Why exactly is this? Does it have something to do with the quadrant that \arctan \frac{0}{x} ends up being in if x is negative?
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  6. #6
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    Quote Originally Posted by TheGreenLaser View Post
    Interesting. Why exactly is this? Does it have something to do with the quadrant that \arctan \frac{0}{x} ends up being in if x is negative?
    Here is the rule for arguments.
    Arg(z) = \left\{ {\begin{array}{*{20}c}<br />
   {\arctan \left( {\frac{y}<br />
{x}} \right),} & {x > 0}  \\<br />
   {\arctan \left( {\frac{y}<br />
{x}} \right) + \pi ,} & {x < 0\;\& \,y \geqslant 0}  \\<br />
   {\arctan \left( {\frac{y}<br />
{x}} \right) - \pi ,} & {x < 0\;\& \,y < 0}  \\<br /> <br />
 \end{array} } \right..

    In the case x=0 then it is \text{sgn}(y)\frac{\pi}{2}
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  7. #7
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    Not exactly the quadrant because arctan \frac{0}{x} is not in any quadrant- it is on the boundary between two quadrants. But you are correct that \tan(\theta) goes from -\infty to \infty for -\pi/2\le \theta\le \pi/2- and then repeats for \pi/2\le\theta\le 3\pi/2. And that means that tangent does not have a "true" inverse for 0\le\theta\le 2\pi. The "principal value" is always between -\pi/2 and \pi/2. To find values between 0 and 2\pi we have to look at the x and y components separately. (When you divide y/x you lose information about the signs of y and x separately.)

    I think it is better to look at it purely geometrically. The complex numbers can be thought of as points in the "complex" plane. The complex number a+ bi corresponds to the point with coordinates (a, b). That is, the "x" axis is the "real axis" and the "y" axis is the imaginary axis. The "argument" of the complex number a+ bi is the angle the line from (a, b) to the origin makes with the positive x-axis. For any positive real number, that angle is 0. For any imaginary number, ai, with a positive, that angle is \pi/2. For any negative real number, that angle is \pi and for any imaginary number, ai, with a negative, that angle is 3\pi/2.

    Yet another way of looking at it: a+ bi= r e^{i\theta}= r(cos(\theta)+ i sin(\theta)) with r always non-negative. In particular, if \theta= 0, a+ bi= r e^{i(0)}= r(cos(0)+ i sin(0))= r(1+ i0)= r which is non-negative. If \theta= \pi, a+ bi= r e^{i\pi}= r(cos(\pi)+ i sin(\pi))= r(-1+ i(0))= -r which is negative.

    Also, if \theta= \pi/2, a+ bi= r (cos(\pi/2}+ isin(\pi/2))= r(0+ i)= ri, a "positive" imaginary number while if \theta= 3\pi/2 (or -\pi/2, then a+ bi= r(cos(3\pi/2)+ i sin(3\pi/2))= r(cos(-\pi/2)+i sin(-\pi/2))= r(0- i)= -ri, a "negative" imaginary number.
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  8. #8
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    Thanks to both of you for the help. More and more I'm realizing that I should have done a better job of understanding trigonometry in highschool. It seems to just be popping up everywhere. I got high 90's in math, but I never really spent time to truly understand trig. I mean I understood it better than most people, but for the most part I just got really good at following the rules, the consequence being that I very quickly forgot after graduation.
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  9. #9
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    Quote Originally Posted by TheGreenLaser View Post
    How did you get pi as the argument? We were taught that if you have a complex number z, then
    z = x + yi = r \exp(i\theta) where r = \sqrt{x^2 + y^2} and \theta = \arctan \frac{y}{x}

    So in this case wouldn't \theta = \arctan \frac{0}{-64} = \arctan 0 = 0 ?
    Do you understand the geometric meaning of argument? Plot z on an argand diagram. The argument is now obvious.
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    Quote Originally Posted by mr fantastic View Post
    Do you understand the geometric meaning of argument? Plot z on an argand diagram. The argument is now obvious.
    No, I can't say I do fully understand the geometric meaning of an argument. Do you mind explaining?
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  11. #11
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    Quote Originally Posted by TheGreenLaser View Post
    No, I can't say I do fully understand the geometric meaning of an argument. Do you mind explaining?
    It will be explained in your class notes or textbook. Google will find numerous websites that explain. You cannot understand the polar form of a complex number without understanding this concept (in my opinion, anyway).
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  12. #12
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    Alright, thanks.
    Last edited by mr fantastic; November 7th 2010 at 02:03 PM. Reason: Moved new question to new thread.
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