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!
If we let $\displaystyle z=a+bi$ as a point (a,b) in the complex plane, and we let r and $\displaystyle {\theta}$ be polar coordinates of that point, then:
$\displaystyle a=rcos{\theta}, \;\ b=rsin{\theta}$
So z can be expressed as $\displaystyle z=a+bi=(rcos{\theta})+(rsin{\theta})i=
r(cos{\theta}+isin{\theta})$
$\displaystyle r=\sqrt{a^{2}+b^{2}}, \;\ tan^{-1}(\frac{b}{a})$
I assume this is what they're wanting.
Plot the point (-3,2). Find the distance from origin to that point. That is r.
Because the point is in quadrant II $\displaystyle \Theta = \pi + \arctan \left( {\frac{2}{{ - 3}}} \right)$.
That is the angle made with the positive x-axis, in other word the argument of the number.
Now we have $\displaystyle - 3 + 2i = r\left( {\cos (\Theta ) + i\sin (\Theta )} \right)$
Not quite right.
Here is the rule that can be programmed into most calculators.
$\displaystyle \arg (x + yi) = \left\{ {\begin{array}{lc}
{\arctan \left( {\frac{y}{x}} \right)} & {x > 0} \\
{\arctan \left( {\frac{y}{x}} \right) + \pi } & {x < 0\;\& \;y > 0} \\
{\arctan \left( {\frac{y}{x}} \right) - \pi } & {x < 0\;\& \;y < 0} \\
\end{array}} \right.$