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Math Help - Solving Complex Numbers.

  1. #1
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    Solving Complex Numbers.

    Find all complex numbers z, which satisfy z^(2)= i - 1.

    I have come across this question during my tutorial and I cant seem to solve it. I need some reference. Thank you---> to the kind soul who is able to help me out here... =D

    God Bless.
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    \displaystyle |z|=\sqrt{2} \ arg(z)=tan^{-1}\frac{y}{x}=tan^{-1}(-1)=\frac{3\pi}{4}

    \displaystyle k=0,1 \ n=2 \ w_k=r^{1/n}*\left[cos\left(\frac{\theta+2\pi k}{n}\right)+isin\left(\frac{\theta+2\pi k}{n}\right)\right]

    \displaysyle w_0=\sqrt{2}^{1/2}*\left[cos\left(\frac{\theta+2\pi *0}{2}\right)+isin\left(\frac{\theta+2\pi *0}{2}\right)\right]\cdots

    \displaystyle w_1=\cdots
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  3. #3
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    Quote Originally Posted by saykhong View Post
    Find all complex numbers z, which satisfy z^(2)= i - 1.

    I have come across this question during my tutorial and I cant seem to solve it. I need some reference. Thank you---> to the kind soul who is able to help me out here... =D

    God Bless.
    Exponential equations are easiest to solve when you put the complex numbers in exponential form. Remember that the are always 2 square roots, 3 cube roots, 4 fourth roots, etc, and they are all evenly spaced around a circle.

    So for square roots, they are separated by an angle of \displaystyle \frac{2\pi}{2} = \pi.


    Anyway, \displaystyle |-1 + i| = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}.

    \displaystyle \arg{(-1 + i)} = \pi - \arctan{\frac{1}{1}} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.


    So that means \displaystyle -1 + i = \sqrt{2}\,e^{\frac{3\pi i}{4}}.


    Therefore, \displaystyle z^2 = \sqrt{2}\,e^{\frac{3\pi}{4}}

    \displaystyle z = \left(\sqrt{2}\,e^{\frac{3\pi}{4}}\right)^{\frac{1  }{2}}

    \displaystyle z = \sqrt[4]{2}\, e^{\frac{3\pi}{8}}.


    That is the first square root. The second has an extra \displaystyle \pi added to its angle. So the other square root is

    \displaystyle z = \sqrt[4]{2}\,e^{\frac{11\pi}{8}}.
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    That was really some good help. And please keep the spirit up high in this forum! Its because of people like that people like us continue to learn everyday!
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