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    Question:

    Let z = a + bi where a, b are real numbers. If z/z* = c + di, where c,d are real numbers, prove that c^2 + d^2 = 1
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    (Original post by vladtheimpaler)
    Question:

    Let z = a + bi where a, b are real numbers. If z/z* = c + di, where c,d are real numbers, prove that c^2 + d^2 = 1

    I've got to a = ca - db and b = da - bc but I have no idea what to do next...
    Assuming you're not using the modulus function.

    So you have z/z*, which is (a+ib)/(a-ib), and you need to get this into the form c + di. Standard technique is to multiply top and bottom by the conjugate of the denominator. Which will give you values for c, and d.
    Then just check that the sum of their squares is 1.

    Enough?
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    Thanks, I see where I've gone wrong, but I still can't get the answer.... any more help?
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    (Original post by vladtheimpaler)
    Thanks, I see where I've gone wrong, but I still can't get the answer.... any more help?
    So what did you get for c and d?
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    \frac{a+bi}{a-bi}=\frac{(a+bi)^2}{(a-bi)(a+bi)}=\frac{a^2+2abi+b^2i^2  }{a^2-b^2i^2}=\frac{a^2+2abi-b^2}{a^2+b^2}=\frac{a^2-b^2}{a^2+b^2}+\frac{2ab}{a^2+b^2  }i

    now you have a real part and an imaginary part, so c = the real part and d = the imaginary part
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    (Original post by Pheylan)
    \frac{a+bi}{a-bi}=\frac{(a+bi)^2}{(a-bi)(a+bi)}=\frac{a^2+2abi+b^2i^2  }{a^2-b^2i^2}=\frac{a^2+2abi-b^2}{a^2+b^2}=\frac{a^2-b^2}{a^2+b^2}+\frac{2ab}{a^2+b^2  }i

    now you have a real part and an imaginary part, so c = the real part and d = the imaginary part
    Thanks again, this is exactly where I got to but I still don't see how I do the last bit; making c^2 + d^2 = 1
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    (Original post by vladtheimpaler)
    Thanks again, this is exactly where I got to but I still don't see how I do the last bit; making c^2 + d^2 = 1
    \frac{(a^2-b^2)(a^2-b^2)}{(a^2+b^2)(a^2+b^2)}+\frac{  4a^2b^2}{(a^2+b^2)(a^2+b^2)}

    \frac{a^4-2a^2b^2+b^4+4a^2b^2}{a^4+2a^2b^2  +b^4}

    \frac{a^4+b^4+2a^2b^2}{a^4+b^4+2  a^2b^2}

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