Represent the complex number graphically, and find the trigonometric form of the number
-7+4i
You can only plot $\displaystyle -7+3i$ on a complex plane, if you look at the article, it also shows you how you go about finding to polar form (which involves trig).
You should be able to work it from there, but if you need help, here is the worked solution for converting it to polar form:
Find the argument (aka the magnitude or the size of the line):
$\displaystyle |-7+3i|= \sqrt{(-7)^2 + 3^2}$
$\displaystyle |-7+3i|= \sqrt{58}$
Find the angle with the horizontal:
$\displaystyle \tan{\theta} = \frac{3}{-7}$
$\displaystyle \theta = \arctan{\frac{3}{-7}}$
$\displaystyle \theta \approx -0.404 (\approx -23.20 \mbox{ degrees})$
We know the point is in the second quadrant, therefore:
$\displaystyle \theta \approx 2.737 (\approx 156.80 \mbox{ degrees})$
This finally leaves us with:
$\displaystyle -7+3i = \sqrt{58} \mbox{cis }2.737$
$\displaystyle -7+3i = \sqrt{58} (\cos{2.737} + i \sin{2.737}$
Well, I added a worked solution. As for plotting the dot. It's quite easy:
Look at the complex plane as being just like any other plane with an X and a Y axis, except that instead of being X and Y, they are $\displaystyle \mbox{Im}(Z)$ (the Imaginary component) and $\displaystyle \mbox{Re}(Z)$ (the Real component) respectively.
In your case with $\displaystyle -7+3i$, $\displaystyle \mbox{Im}(Z) = 3$ and $\displaystyle \mbox{Re}(Z) = -7$. So this in effect would be just like plotting the point (-7,3) on a standard plane.
let $\displaystyle z=-7+4i
$
let $\displaystyle z'=7+4i$
$\displaystyle
\tan\theta_1=\frac{4}{7}$
$\displaystyle \theta_1=\tan^{-1}\frac{4}{7}$
as z lies in IInd quadrant , $\displaystyle \arg(z)=\theta=\pi-\theta_1=\pi-0.18$
$\displaystyle r=\sqrt{7^2+4^2}=\sqrt{65}$
now $\displaystyle z=re^{i\theta}=r(\cos(\pi-0.18)+i\sin(\pi-0.18))=\sqrt{65}(i\sin 0.18-\cos 0.18)$