Ihave more questions so if you could with those after answering this one worth 30 pts that would be wonderful!

1.
express the complex number in trigonometric form. -2

2.
find the product of z1 and z2, where z1 = 8(cos 40° + i sin 40°) and z2 = 4(cos 135° + i sin 135°).

3.4.5.6. in a message\\

message me after so we might be able to do the others? if you find its worth your while to me! and i can try to you with your

Problem 1) 

2*(cos(180)+i*sin(180))



Rewrite -2 as -2+0i. Plotting this complex number on the complex plane will lead us to the point (-2,0). The x coordinate is the real part; the y coordinate is the imaginary part.

The distance from (0,0) to (-2,0) is 2 units, so r = 2
The angle is 180 degrees so theta = 180

This is why the answer is 2*(cos(180)+i*sin(180))

Since the general form is r*(cos(theta)+i*sin(theta))



Problem 2)

32*(cos(175)+i*sin(175))



Let's say we had two general complex numbers z1 and z2 defined as such
z1 = r1*(cos(theta1)+i*sin(theta1))
z2 = r2*(cos(theta2)+i*sin(theta2))

Multiplying z1 and z1 will have us multiply the r values (r1 and r2) and adding the theta values (theta1 and theta2)

So this means
z1*z2 = (r1*r2)*(cos(theta1+theta2)+i*sin(theta1+theta2))

Let's use that rule with the given z1 and z2 values
z1*z2 = (r1*r2)*(cos(theta1+theta2)+i*sin(theta1+theta2))
z1*z2 = (8*4)*(cos(40+135)+i*sin(40+135))
z1*z2 = 32*(cos(175)+i*sin(175))

4should be the last one n=4

Two because two of one thing is two.