Question

Trig question
When doing Polar Form of a Complex Number, if the point plotted is in quadrant 2 or 3, do you add pi?

Answer 1

@jim_thompson5910 please help

Answer 2

what is the point?

Answer 3

4/-3

Answer 4

(4, -3) ?

Answer 5

yeah its in the form z = a + bi (-3 + 4i)

Answer 6

(4,-3) = (a,b)

so a+bi = 4-3i

Answer 7

assuming (4,-3) is in the form (a,b) where a+bi

Answer 8

my bad. In this case it's -3 + 4i

Answer 9

is the initial point (4,-3) or (-3,4)

Answer 10

(-3,4)

Answer 11

and you want to convert this to polar form right?

Answer 12

yes

Answer 13

one may note and angle in quadrant II woudln't amount to a \(\pi\) unles is \(\pi\) itself

Answer 14

(x,y) = (-3,4)

x = -3
y = 4

now use the formula

\[\Large r = \sqrt{x^2+y^2}\]

to find the value of r

Answer 15

so r = 5

Answer 16

yes r = 5

Answer 17

to find theta, you'll use

\[\Large \theta = \arctan\left(\frac{y}{x}\right)\]

you need to add on pi (I think this is what you were initially asking about) because you will start off in Q4, but adding pi gets you to Q2 where the angle is

Answer 18

so again, the arctan function in this case will return an angle in Q4

adding pi to that result brings the angle to Q2

Answer 19

YES that is what I was confused about, when you add pi to theta

Answer 20

so hopefully that clears things up?

Answer 21

so you have to add pi when its in Q1 or Q4?

Answer 22

The range of arctan is -pi/2 < y < pi/2

so that means that the outputs of arctan are either in Q1 or Q4

Answer 23

if you get something in Q4, but the angle is really in Q2, you have to add pi

but if the angle is really in Q4, then you just leave it as it is

Answer 24

same for Q1 and Q3

Answer 25

ok.. thank you :)

Answer 26

np