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Thread: How to Computing Square root of a Complex number...

  1. #1
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    How to Computing Square root of a Complex number...

    Hi,

    How do I compute square root of a complexed number without a calculator?
    In other words how should I think and break down the following with just pen and paper?

    $\displaystyle \displaystyle\sqrt{8i}$

    I would appriciate any guidance.
    Thank you
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  2. #2
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    Convert it to exponential form.

    $\displaystyle \displaystyle \sqrt{8i} = \sqrt{8e^{\frac{\pi i}{2}}}$

    $\displaystyle \displaystyle = 2\sqrt{2}(e^{\frac{\pi i}{2}})^{\frac{1}{2}}$

    $\displaystyle \displaystyle = 2\sqrt{2}e^{\frac{\pi i}{4}}$

    $\displaystyle \displaystyle = 2\sqrt{2}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)$

    $\displaystyle \displaystyle = 2\sqrt{2}\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}i}{2}\right)$

    $\displaystyle \displaystyle = 2 + 2i$.


    Note that this only gives one of the square roots, but you can find the other using the fact that there are always 2 square roots and they are evenly spaced around a circle.
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  3. #3
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    Hello, 4Math!

    Here is a primitive method . . .


    Find: .$\displaystyle \sqrt{8i}$

    Let: $\displaystyle a + bi \:=\:\sqrt{8i}$ . where $\displaystyle \,a$ and $\displaystyle \,b$ are real.

    Square both sides: .$\displaystyle (a + bi)^2\;=\;(\sqrt{8i})^2 $

    And we have: .$\displaystyle (a^2-b^2) + (2ab)i \;=\;8i$


    Equate real and imaginary components: .$\displaystyle \begin{array}{cccc}a^2-b^2 &=& 0 & [1] \\ 2ab &=& 8 & [2]\end{array}$

    From [2] we have: .$\displaystyle b \:=\:\frac{4}{a}\;\;[3]$

    Substitute into [1]: .$\displaystyle a^2 - \left(\tfrac{4}{a}\right)^2 \:=\:0 \quad\Rightarrow\quad a^4 \:=\:16$

    . . Hence: .$\displaystyle a \:=\:\pm2$

    Substitute into [3]: .$\displaystyle b \:=\:\dfrac{4}{\pm2} \:=\:\pm2$


    Hence: .$\displaystyle a + bi \;=\;\pm2 \pm 2i$


    Therefore: .$\displaystyle \sqrt{8i} \;=\;\pm2(1 + i)$
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