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Thread: complex numbers

  1. #1
    Junior Member
    Jun 2009

    complex numbers

    what are the roots of square root of 1-1.73j giving results in the rectangular and in exponential form
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  2. #2
    Senior Member Twig's Avatar
    Mar 2008

    I am not sure exactly what you mean, but if  z \in \mathbb{C} \; z=a+bi , then  |z| = \sqrt{a^{2}+b^{2}}

    And the argument of the complex number would be:

     2\pi - arctan(1.73) , since you are in the fourth quadrant. Do you understand why here?

    A complex number can be expressed:  |z|(cos(\theta)+i\cdot sin(\theta) where  \theta is the argument of z, or equivalently  e^{ln(|z|)}(cos(\theta)+i \cdot sin(\theta)
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  3. #3
    MHF Contributor

    Apr 2005
    Is that really 1.73 or is it \sqrt{3}? I am going to assume you mean \sqrt{3}.

    Once you have z= 1- i\sqrt{3} in polar form, z= r(cos(\theta)+ i sin(\theta)), which is what Twig gave you, its square roots are r^{1/2}(cos(\theta/2)+ i sin(\theta/2) and r^{1/2}(cos(\theta/2+ \pi)+ i sin(\theta/2+ \pi).

    Here, r^{1/2} is the "principle" root of the real number r: for this problem r= \sqrt{1+ (\sqrt{3})^2}= 2 and \theta= tan^{-1}(-\sqrt{3})= -\pi/3. (Twig's value was 2\pi- \pi/3= 5\pi/3 but I prefer -\pi/3. They give the same z.)

    So the square roots of 1- i\sqrt{3} are \sqrt{2}(cos(-\pi/6)+ i sin(-\pi/6))= \sqrt{2}(\frac{1}{2}- \frac{\sqrt{3}}{2}i) and \sqrt{2}(cos(5\pi/6)+ i sin(5\pi/6))= \sqrt{2}(-\frac{1}{2}+ \frac{\sqrt{3}}{2}i).

    (Sorry, but I just can't bring myself to write "j" instead of "i"! Oh, those engineers and their jmaginary numbers!)
    Last edited by mr fantastic; Jun 12th 2009 at 04:13 AM. Reason: Fixed some latex tags.
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