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Math Help - Complex number-finding roots

  1. #1
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    Complex number-finding roots

    Find the roots of the equation z^2=21-20i
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  2. #2
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    Hello, kingkaisai2!

    Edit: I corrected my wrong sign below ... Sorry for any confusion it caused.

    I don't know what methods you've been taught . . .


    Find the roots of the equation z^2 \:=\:21-20i

    Let z = a + bi, where a and b are real.

    We have: . (a + bi) \:= \:21 - 20i

    Then: . a^2 + 2abi+  b^2i^2\:=\: 21 - 20i\quad\Rightarrow\quad (a^2 - b^2) + 2abi \:=\:21 - 20i

    Equate real and imaginary components: . \begin{array}{cc}a^2 - b^2 \:= \:21\\ 2ab \:= \:-20\end{array} \begin{array}{cc}(1)\\(2)\end{array}

    From (2), we have: . b = -\frac{10}{a}

    Substitute into (1): . a^2 - \left(-\frac{10}{a}\right)^2\:=\:21\quad\Rightarrow\quad a^2 - \frac{100}{a^2}\:=\:21

    Multiply by a^2:\;\;a^4 - 100\:=\:21a^2\quad\Rightarrow\quad a^4 - 21a^2 - 100\:=\:0

    Factor: . (a^2 + 4)(a^2 - 25)\:=\:0


    We have two equations to solve:

    . . a^2 + 4\:=\:0\quad\Rightarrow\quad a^2\,=\,-4\;\cdots \text{ but }a\text{ must be }real.

    . . a^2 - 25\:=\:0\quad\Rightarrow\quad a^2\,= \,25\quad\Rightarrow\quad a\,=\,\pm5

    Substitute into (2): . b \:= \:-\frac{10}{\pm5} \:=\:\mp2


    Therefore: . \boxed{z\;=\;\{5 - 2i,\;-5 + 2i\}}

    Last edited by Soroban; July 31st 2006 at 03:06 PM.
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  3. #3
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    Quote Originally Posted by Soroban
    Hello, kingkaisai2!

    I don't know what methods you've been taught . . .



    Let z = a + bi, where a and b are real.

    We have: . (a + bi) \:= \:21 - 20i

    Then: . a^2 + 2abi - b^2i^2\:=\: 21 - 20i\quad\Rightarrow\quad (a^2 - b^2) + 2abi \:=\:21 - 20i
    shouldn't it be (a^2+b^2)+2abi=21-20i? because i^2 equals -1
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