Question

principal value of the argument
-7/61+16/61(i)
in the range of -pi to pi
help anyone?

Answer 1

solve
\[\tan(\theta)=\frac{b}{a}\]

Answer 2

\[a=\frac{-7}{61},b=\frac{16}{61}\]
\[\frac{b}{a}=-\frac{16}{7}\]so
\[\theta=\tan^{-1}(-\frac{16}{7})\]

Answer 3

now you need a calculator

Answer 4

in the answer it says pi - tan^-1 (16/7)

Answer 5

that's what confused me

Answer 6

well if you don't mind me saying that is a really really stupid answer that tells you nothing

Answer 7

analysis IS STUPID i have an exam tomorrow haha

Answer 8

just tells you that
\[\pi-\tan^{-1}(\frac{16}{7})=\tan^{-1}(-\frac{16}{7})\]

Answer 9

that's not the same is it?

Answer 10

inverse tangent already has range
\[[-\frac{\pi}{2},\frac{\pi}{2}]\]

Answer 11

i am sticking with my answer.

Answer 12

actually now that i look at it i don't even understand the question "has range -pi to pi"

Answer 13

-pi to pi is the whole cirlce

Answer 14

http://www.wolframalpha.com/input/?i=-7%2F61%2B16%2F61%28i%29

Answer 15

2 pi is a circle

Answer 16

oooooooooooooooh i see
you are in quadrant 2

Answer 17

it says the principal value of the argument is in the range (-pi,pi]

Answer 18

got it. ok fine

Answer 19

you are in quadrant 2
so angle is between \[\frac{\pi}{2}\] and \[\pi\]

Answer 20

add pi to
\[\tan^{-1}(-\frac{16}{7})\]

Answer 21

thanks, how would i know when to add pi?