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Math Help - Imaginary numbers and their polar form

  1. #1
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    Imaginary numbers and their polar form

    Hey guys, for my Complex Analysis exam coming up, my lecturer has said that no calculators will be allowed. This is fine, but I've been having trouble switching between imaginary numbers' standard and polar forms. I can find the polar forms of numbers that have even denominators in their polar for (i.e. i = e^(pi/2)i), but I don't know how to find the polar form of numbers with odd numerators (i.e. -1/2 + (sqrt(3)/2)i = e^(2pi/3)i). If anyone could help me with this, or has an online resource that I can look at, I'd greatly appreciate it.
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    Switching from Cartesian to Polar form is easy, as long as you know which quadrant the complex number lies in.

    For a complex number \displaystyle z = x + iy, we can convert it to the polar form \displaystyle z = r\,e^{i\theta} using

    \displaystyle r = \sqrt{x^2 + y^2}

    and, depending on the quadrant of the complex number

    Quadrant 1: \displaystyle \theta = \arctan{\left|\frac{y}{x}\right|}

    Quadrant 2: \displaystyle \theta = \pi - \arctan{\left|\frac{y}{x}\right|}

    Quadrant 3: \displaystyle \theta = -\pi + \arctan{\left|\frac{y}{x}\right|}

    Quadrant 4: \displaystyle \theta = -\arctan{\left|\frac{y}{x}\right|}
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    Surely I would need a calculator to compute arctan?
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    Quote Originally Posted by Alexrey View Post
    Surely I would need a calculator to compute arctan?
    If you are studying complex numbers, you surely must know the exact trigonometric values from special triangles, for example \displaystyle \tan{\left(\frac{\pi}{6}\right)} = \frac{1}{\sqrt{3}}. I would expect that you would only be given angles you can evaluate the arctangent of from special triangles.
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    Quote Originally Posted by Alexrey View Post
    Surely I would need a calculator to compute arctan?
    Without a calculator you can only deal with angles with basic measure:
    0,~\frac{\pi}{6},~\frac{\pi}{4}, ~\frac{\pi}{3},~\frac{\pi}2},~\&~\pi.
    So just learn those special angles.
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    Unfortunately I've never been taught the special angles, I'll check them out though. Thanks guys
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