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Math Help - Complex number equation

  1. #1
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    Complex number equation

    Hi;

    Find the square roots of 5 + 12i. Hence find the roots of the equation
    z^2-(1+6i)z-10=0

    Thanks for helping up.
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  2. #2
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    Hello, eureka!

    Didn't I do one of these for you already?


    Find the square roots of 5 + 12i.
    Hence find the roots of the equation: \,z^2 - (1+6i)z - 10\:=\:0

    Let z \:=\:5 + 12i

    In polar form: . z\:=\:13(\cos\theta + i\sin\theta) . . . where \tan\theta = \frac{12}{5}

    Then: . z^{\frac{1}{2}}\;=\;13^{\frac{1}{2}}\left[\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right] [1]


    Since \tan\theta = \frac{12}{5}, then: \cos\theta = \frac{5}{13}

    . . \cos\frac{\theta}{2}\:=\:\sqrt{\frac{1 + \cos\theta}{2}} \:=\:\sqrt{\frac{1 + \frac{5}{13}}{2}} \:=\: \sqrt{\frac{18}{26}}\:=\:\frac{3}{\sqrt{13}}

    . . \sin\frac{\theta}{2}\:=\:\sqrt{\frac{1 - \cos\theta}{2}} \:=\:\sqrt{\frac{1 - \frac{5}{13}}{2}} \:=\:\sqrt{\frac{8}{26}} \:=\:\frac{2}{\sqrt{13}}


    Substitute into [1]: . z^{\frac{1}{2}} \;=\;\sqrt{13}\left(\frac{3}{\sqrt{13}} + i\frac{2}{\sqrt{13}}\right) \;=\;3 + 2i


    Therefore: . \sqrt{5 + 12i} \;=\;\boxed{\pm(3 + 2i)}

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