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

**coopsterdude** · August 14th 2007, 04:08 PM

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!

**galactus** · August 14th 2007, 04:25 PM

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 $z=a+bi=(rcos{\theta})+(rsin{\theta})i= r(cos{\theta}+isin{\theta})$

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

I assume this is what they're wanting.

**Plato** · August 14th 2007, 04:34 PM

Originally Posted by coopsterdude

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 $- 3 + 2i = r\left( {\cos (\Theta ) + i\sin (\Theta )} \right)$

**Plato** · August 14th 2007, 04:45 PM

Originally Posted by galactus

The angle is called the argument
${\theta}=arg (z)$

Not quite right.
Here is the rule that can be programmed into most calculators.
$\arg (x + yi) = \left\{ {\begin{array}{lc} {\arctan \left( {\frac{y}{x}} \right)} & {x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi } & {x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi } & {x < 0\;\& \;y < 0} \\ \end{array}} \right.$

**coopsterdude** · August 14th 2007, 10:52 PM

Thanks guys, that explains a lot. I think I can now use the calculator to work with the complex numbers.

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