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Thread: 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 $\displaystyle z^2 \:=\:21-20i$

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

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

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

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

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

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

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

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


    We have two equations to solve:

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

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

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


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

    Last edited by Soroban; Jul 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 $\displaystyle z = a + bi$, where $\displaystyle a$ and $\displaystyle b$ are real.

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

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