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write -4 as -4+0i
that is in the form a+bi

Ok.

a = -4
b = 0
plug those values into the formulas
r = sqrt(a^2 + b^2)
theta = arctan(b/a)

So r = 4 and theta = 0?

r = 4 is correct
but theta = 0 is not

|dw:1437696080574:dw|

Isn't 0/-4 equal to 0, and the arctan(0) just 0?

So is r = 4 and theta = -4...?

theta = 180 degrees = pi radians

so theta = pi ?

yeah

Ok so let me try and see if I can get the answer.

you are correct

Thank you! I'll Medal you (:

you're welcome

Could you help me do the same thing for -6i?

-6i = 0 + (-6i)
now a = 0 and b = -6

use the formulas
r = sqrt(a^2 + b^2)
theta = arctan(b/a)

draw a graph to make sure you get the theta in the right spot

So would theta = pi?

what is the value of b/a this time?

Undefined actually

where is tan(theta) undefined?

Isn't it at 90 degrees and 270 degrees?

correct

|dw:1437697203957:dw|

|dw:1437697245770:dw|

So theta = 3pi/2

yep

Therefore making the answer:
6(cos270degrees + i sin270degrees)

correct

what's the question

3-3i

that's not an equation. An equation needs an = sign

Sorry, meant expression.

compare a+bi with 3-3i to see that a = 3 and b = -3

theta is not -1

b/a = -1, that is true

arctan(b/a) = arctan(-1) = ??

-45...

So 315 degrees?

yes

So would the answer be:
3SqRt2(cos7pi/4 + i sin7pi/4) ?

correct

that would be the radian form

you're welcome

Yes, that is what the answers are listed as.