Question 1201677
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Edit: Sorry I had a brainfart. I mentioned the angle in Q2 when it should have been Q4.
I have fixed the mistake.


I'll do problem 1 to get you started.


2-2i = 2 + (-2)i 
this is written in the form a+bi
a = 2
b = -2


r = modulus
r = sqrt(a^2+b^2)
r = sqrt((2)^2+(-2)^2)
r = sqrt(4+4)
r = sqrt(8)
r = sqrt(4*2)
r = sqrt(4)*sqrt(2)
r = 2*sqrt(2)


theta = argument
theta = arctan(b/a)
theta = arctan(-2/2)
theta = arctan(-1)
theta = -45°
Add 360 degrees to find a coterminal angle in the range 0 ≤ theta < 360
-45+360 = 315



Summary:
r = 2*sqrt(2)
theta = 315°


The polar point (r, theta) is <font color=red>(2*sqrt(2), 315°)</font>


Another way you could express the answer would be to say the following
z = r*( cos(theta) + i*sin(theta) )
z = <font color=red>2*sqrt(2)*( cos(315°) + i*sin(315°) )</font>
z = <font color=red>2*sqrt(2)*cis(315°)</font>
where "cis" is shorthand for "cosine i sine"
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