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Math Help - How can I use my graphing calculator to solve this complex number problem?

  1. #1
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    How can I use my graphing calculator to solve this complex number problem?

    The question in my book says use a graphing calculator to represent -3+2i in trigonometric form.

    I'm using a TI-83 but cannot solve the problem. Please help!
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  2. #2
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    If we let z=a+bi as a point (a,b) in the complex plane, and we let r and {\theta} be polar coordinates of that point, then:

    a=rcos{\theta}, \;\ b=rsin{\theta}

    So z can be expressed as z=a+bi=(rcos{\theta})+(rsin{\theta})i=<br />
r(cos{\theta}+isin{\theta})

    r=\sqrt{a^{2}+b^{2}}, \;\ tan^{-1}(\frac{b}{a})

    I assume this is what they're wanting.
    Last edited by galactus; August 15th 2007 at 09:40 AM.
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  3. #3
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    Quote Originally Posted by coopsterdude View Post
    The question in my book says use a graphing calculator to represent -3+2i in trigonometric form. I'm using a TI-83 but cannot solve the problem.
    Plot the point (-3,2). Find the distance from origin to that point. That is r.
    Because the point is in quadrant II  \Theta  = \pi  + \arctan \left( {\frac{2}{{ - 3}}} \right).
    That is the angle made with the positive x-axis, in other word the argument of the number.

    Now we have   - 3 + 2i = r\left( {\cos (\Theta ) + i\sin (\Theta )} \right)
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  4. #4
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    Quote Originally Posted by galactus View Post
    The angle is called the argument
    {\theta}=arg (z)
    Not quite right.
    Here is the rule that can be programmed into most calculators.
    \arg (x + yi) = \left\{ {\begin{array}{lc}<br />
   {\arctan \left( {\frac{y}{x}} \right)} & {x > 0}  \\<br />
   {\arctan \left( {\frac{y}{x}} \right) + \pi } & {x < 0\;\& \;y > 0}  \\<br />
   {\arctan \left( {\frac{y}{x}} \right) - \pi } & {x < 0\;\& \;y < 0}  \\<br />
\end{array}} \right.
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  5. #5
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    Thanks guys, that explains a lot. I think I can now use the calculator to work with the complex numbers.
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