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Math Help - Square roots of complex numbers.

  1. #1
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    As part of a question I have to find the square root of -24 + 10i. I know that 1 + 5i is one square root, but I have no idea how to find this other than guess and check. I tried converting to polar form, but yet again it's the argument that's getting me. I know that x = -24, y = 10, and r = 26. But there doesn't seem to be a "nice" argument based on those values. arctan(10/-24) = 157.380... degrees, arccos(-24/26) = 157.380... degrees, and arcsin(10/26) = 157.380... degrees. So it's consistent, but it seems like there's no way it's going to give me an exact value for the square root, even though I know that there's a 'nice' root (1 + 5i). I thought about maybe just leaving the argument as (arccos(-24/26)) and then manipulating double angle identities, but I wasn't sure how to properly do this. Any help?
    Last edited by mr fantastic; November 7th 2010 at 02:02 PM. Reason: Moved from another thread.
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    In this the answer is simple: (-1-5i).
    Square roots occur as negatives,
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    I guess my question wasn't clear enough. I understand that (-1-5i) would also be a square root, I just don't understand how to find +/-(1+5i) in the first place other than guess and check. The method I was taught was to convert everything into polar form, but (-24+10i) doesn't seem to convert nicely, and decimal answers won't do, I need exact answers.
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    Quote Originally Posted by TheGreenLaser View Post
    I guess my question wasn't clear enough. I understand that (-1-5i) would also be a square root, I just don't understand how to find +/-(1+5i) in the first place other than guess and check. The method I was taught was to convert everything into polar form, but (-24+10i) doesn't seem to convert nicely, and decimal answers won't do, I need exact answers.
    The easiest approach here is to let the square root be a + ib where a and b are real. Then:

    -24 + 10 i = (a + ib)^2 = (a^2 - b^2) + i(2ab).

    Equate real and imaginary parts:

    -24 = a^2 - b^2 .... (1)

    10 = 2ab \Rightarrow 5 = ab \Rightarrow a = \frac{5}{b} .... (2)

    Substitute (2) into (1): -24 = \frac{25}{b^2} - b^2 \Rightarrow b^4 - 24b^2 - 25 = 0 \Rightarrow (b^2 - 25)(b^2 + 1) \Rightarrow b^2 = 25 (the other solution is rejected - why?).

    etc.
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    Another approach is to remember that there are 2 square roots and they are evenly spaced around a circle. So the other square root has the same length and differs from the other by an angle of \displaystyle \frac{\pi}{2}.
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    Quote Originally Posted by mr fantastic View Post
    The easiest approach here is to let the square root be a + ib where a and b are real. Then:

    -24 + 10 i = (a + ib)^2 = (a^2 - b^2) + i(2ab).

    Equate real and imaginary parts:

    -24 = a^2 - b^2 .... (1)

    10 = 2ab \Rightarrow 5 = ab \Rightarrow a = \frac{5}{b} .... (2)

    Substitute (2) into (1): -24 = \frac{25}{b^2} - b^2 \Rightarrow b^4 - 24b^2 - 25 = 0 \Rightarrow (b^2 - 25)(b^2 + 1) \Rightarrow b^2 = 25 (the other solution is rejected - why?).

    etc.
    So why is the one solution rejected?
    (b^2 + 1) = 0 \Rightarrow b^2 = -1 \rightarrow b = i, -i

    I've substituted these values back into equation (2) to get a value for a, a = \frac{5}{b} = \frac{5}{i} = \frac{5i}{i^2} = \frac{5i}{-1} = -5i and then used that value to form the complex number a +bi = -5i + (i)i = -5i -1 = -1 - 5i which is one of the roots I'm looking for. Using b = -i just yields 1 + 5i as a root. Is it thrown out because a is supposed to represent the real part, but in that case it becomes the imaginary part? Or is it maybe because it represents the exact same roots as the other?
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  7. #7
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    Quote Originally Posted by TheGreenLaser View Post
    So why is the one solution rejected?
    (b^2 + 1) = 0 \Rightarrow b^2 = -1 \rightarrow b = i, -i

    I've substituted these values back into equation (2) to get a value for a, a = \frac{5}{b} = \frac{5}{i} = \frac{5i}{i^2} = \frac{5i}{-1} = -5i and then used that value to form the complex number a +bi = -5i + (i)i = -5i -1 = -1 - 5i which is one of the roots I'm looking for. Using b = -i just yields 1 + 5i as a root. Is it thrown out because a is supposed to represent the real part, but in that case it becomes the imaginary part? Or is it maybe because it represents the exact same roots as the other?
    Re-read my previous reply:

    snip
    a + ib where a and b are real.
    I will let you think about why a and b are required to be real.

    b = 5 \Rightarrow a = 1, b = -5 \Rightarrow a = -1. Therefore the two roots are 1 + 5i and -1 - 5i.
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