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Math Help - [SOLVED] A complex number

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] A complex number

    Find all the complex numbers such that z^2=1-i.
    My attempt : I wrote z=(a+ib) and with a little bit of some algebra I reached that ab=-\frac{1}{2} which mean that if ab=-\frac{1}{2} then z^2=1-i is satisfied. So the solution is all z such that \forall a,b \in \mathbb{R} / ab=-\frac{1}{2} with z=a+ib. I'm sure there is a nicer way to write the answer. Can you check out my answer and write it in a better way if it is right? Thanks in advance.
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    You are correct that
    2ab=-1

    but remember to equate the real parts as well as the imaginary parts

    so

    (a+ib)^2=1-i

    a^2+2abi-b^2=1-i

    (a^2-b^2)+2abi=1-i

    equating real parts as well gives us a^2-b^2=1

    now solve the two equations simultaneously to find the answer


    as an aside you can always check your answer by plugging a few numbers in, if we try a=1 b=-0.5 which satisfy your condition we have
    (1-\frac{1}{2}i)^2=1-\frac{1}{4}-1i=\frac{3}{4}-1i\neq 1-i

    therefore it cannot be correct on its own
    Last edited by thelostchild; September 14th 2008 at 12:36 PM.
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    MHF Contributor arbolis's Avatar
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    but remember to equate the real parts as well as the imaginary parts

    so







    equating real parts as well gives us
    I didn't forget to do that, the problem is that I didn't reached anything beautiful. Like b=+ or -\sqrt{a^2-1}, or b^4-b^2=\frac{1}{4}, or 4b^2(b^2+1)=1 and so on.
    I'd like to know the answer and then try to reach it.
    I've a similar problem to solve after this so I'll try it alone.
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    Why not just find the two square roots of 1-i?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by arbolis View Post
    I didn't forget to do that, the problem is that I didn't reached anything beautiful. Like b=+ or -\sqrt{a^2-1}, or b^4-b^2=\frac{1}{4}, or 4b^2(b^2+1)=1 and so on.
    I'd like to know the answer and then try to reach it.
    I've a similar problem to solve after this so I'll try it alone.
    you are finding the two square roots of 1 - i

    the answers are a = \pm \sqrt{\frac {1 + \sqrt{2}}2} and b = - \frac 1{2a} ........and of course, you plug in what a is (can you get to that? )

    you could also try converting to polar form, it might be the same, or more work, but not much different in terms of difficulty
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    MHF Contributor arbolis's Avatar
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    I've reached b=\pm \sqrt{\frac{1}{\pm 2\sqrt 2 +2}} so you can imagine to what a is equal. But b is a real number, so I can simply it to \pm \sqrt{\frac{1}{2\sqrt 2 +2}}. I feel I made an error somewhere.
    I've started with a^2-b^2=-2ab.
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    In polar/exponential form 1 - i = \sqrt 2 \left( {\cos \left( {\frac{{ - \pi }}{4}} \right) + i\sin \left( {\frac{{ - \pi }}{4}} \right)} \right) = \sqrt 2 e^{i\left( {\frac{{ - \pi }}{4}} \right)}
    The roots are z_1  = \sqrt[4]{2}\left( {\cos \left( {\frac{{ - \pi }}<br />
{8}} \right) + i\sin \left( {\frac{{ - \pi }}{8}} \right)} \right)\;\& \;z_2  = \sqrt[4]{2}\left( {\cos \left( {\frac{{7\pi }}{8}} \right) + i\sin \left( {\frac{{7\pi }}{8}} \right)} \right)
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  8. #8
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Plato View Post
    In polar/exponential form 1 - i = \sqrt 2 \left( {\cos \left( {\frac{{ - \pi }}{4}} \right) + i\sin \left( {\frac{{ - \pi }}{4}} \right)} \right) = \sqrt 2 e^{i\left( {\frac{{ - \pi }}{4}} \right)}
    The roots are z_1  = \sqrt[4]{2}\left( {\cos \left( {\frac{{ - \pi }}<br />
{8}} \right) + i\sin \left( {\frac{{ - \pi }}{8}} \right)} \right)\;\& \;z_2  = \sqrt[4]{2}\left( {\cos \left( {\frac{{7\pi }}{8}} \right) + i\sin \left( {\frac{{7\pi }}{8}} \right)} \right)
    Ah that's nice, I think I will do that in my exam even if I should simplify -\frac{\pi}{8} and \frac{7\pi}{8}. I think my teacher will give me as much credit as if I solve it without using polar form.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by arbolis View Post
    I've reached b=\pm \sqrt{\frac{1}{\pm 2\sqrt 2 +2}} so you can imagine to what a is equal. But b is a real number, so I can simply it to \pm \sqrt{\frac{1}{2\sqrt 2 +2}}. I feel I made an error somewhere.
    I've started with a^2-b^2=-2ab.
    without doing polar form, you should start with

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

    from (2), b = - \frac 1{2a}. plug that into (1) and solve for a. note that you must take the real answer. then plug that into (2) to solve for b
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