# Exponential form of a complex number

#### blues1

1. The problem statement, all variables and given/known data

if z = -2 + 2i

find r and θ

3. The attempt at a solution

our teacher told us that when we have z = a + bi

r = sqrt(a^2 + b^2)

and θ = tan^-1(b/a)

so here it's supposed to be r = sqrt(8) and θ = - pi/4

but using wolfram alpha to see if the results are matching I get that

sqrt(8)*e^(i*-pi/4) is 2 - 2i

what am I doing wrong? I guess this has to do with the θ, but since it's always tan^-1(b/a) why am I getting different results?

#### HallsofIvy

This is an example of why you should not use formulas without thinking.

b/a and therefore tan(b/a) are exactly the same if you change signs on both a and b. In particular,
$$\frac{-2}{2}= \frac{2}{-2}= -1$$

Also, $tan(\theta)= tan(\pi+ \theta)$. Since tan is "multi-valued", a calculator or computer (or table of trig functions) will get the principle value which is always between $-\pi/2$ and $\pi/2$. To distinguish between the two you need to look at the signs of the individual numbers. (-2, 2) is in the second quadrant so $\theta$ is between $\pi/2$ and $\pi$. In this problem, $\theta= \pi- \pi/4= 3\pi/4$, NOT $-\pi/4$.

#### blues1

This is an example of why you should not use formulas without thinking.

b/a and therefore tan(b/a) are exactly the same if you change signs on both a and b. In particular,
$$\frac{-2}{2}= \frac{2}{-2}= -1$$

Also, $tan(\theta)= tan(\pi+ \theta)$. Since tan is "multi-valued", a calculator or computer (or table of trig functions) will get the principle value which is always between $-\pi/2$ and $\pi/2$. To distinguish between the two you need to look at the signs of the individual numbers. (-2, 2) is in the second quadrant so $\theta$ is between $\pi/2$ and $\pi$. In this problem, $\theta= \pi- \pi/4= 3\pi/4$, NOT $-\pi/4$.
if i remember correct everything is positive from 0 to pi/2, in pi/2 to p only the sin is positive, cos is positive from p to 3pi/2

so if I get a negative θ and z for example is from 0 to pi/2 I should just have tan^-1(b/a)

if I am from pi/2 to p, I should have p - tan^-1(b/a)

from p to 3pi/2 => 3pi/2 - tan^-1(b/a)?

is this how it goes? :/

#### eumyang

Homework Helper
if i remember correct everything is positive from 0 to pi/2, in pi/2 to p only the sin is positive, cos is positive from p to 3pi/2

so if I get a negative θ and z for example is from 0 to pi/2 I should just have tan^-1(b/a)
I don't understand what you are saying. If z = a + bi and both a and b are positive, θ = tan^-1(b/a) would just be positive.

from p to 3pi/2 => 3pi/2 - tan^-1(b/a)?

is this how it goes? :/
No. If z = a + bi and both a and b are negative, then when you find tan^-1(b/a) you will get an angle between 0 and π/2. But you want to be in Quadrant III, so you'll have to add pi.

And you're missing the case of z being in Quadrant IV of the complex coordinate plane.

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