Converting -2(cos π/4 + i sin π/4) to Cartesian, polar and exponential form

[mikecrush]

Homework Statement

how can i convert this : - 2 (cos pai / 4 + i sin pai / 4 ) to Cartesian , Polar and Exponential form ?

Homework Equations

z = ( a + i b)

The Attempt at a Solution

r= -2
tan inverse = pai/4 / pai/4
??

Thank you very much for helping me out

 

[HallsofIvy]

It is already almost in "polar form". If you did not see that immediately, you need to review the definitions. The only reason it is not already in polar form is because the "r" in "$r (cos(\theta)+ i sin(\theta))$" cannot be negative. Draw the line with $\theta= \pi/4$ and go backwards: $-2(cos(\theta)+ i sin(\theta))= 2(cos(\theta+ \pi)+ i sin(\theta+ \pi)$

On thing you should know is that the "r" in a polar
To change to "Cartesian form", just evaluate the functions. What is $cos(\pi/4)$? What is $sin(-\pi/4)$? What are $-2 cos(\pi/4)$ and $-2 sin(\pi/4)$?

The "exponential form" of $r(cos(\theta)+ i sin(\theta))$ is $r e^{i\theta}$. Again, r cannnot be negative.