Question

PLEASE HELP!!! (sqrt 3 - i) in exponetial form

Answer 1

\[\sqrt{3}-i\]

Answer 2

you need two numbers \(r\) and \(\theta\) to write
\[\sqrt3-i=r(\cos(\theta)+i\sin(\theta))\] namely \(r\) and \(\theta\)

Answer 3

i'm so lost. can you help more?

Answer 4

sure
\(r\) is the easiest to find
it is \(|a+bi|=\sqrt{a^2+b^2}\) which in our case is
\[\sqrt{\sqrt3^2+1^2}=\sqrt{3+1}=\sqrt4=2\]

Answer 5

then you need \(\theta\)

Answer 6

\[\cos(\theta)=\frac{a}{4}=\frac{\sqrt3}{2}\]
\[\sin(\theta)=\frac{b}{r}=-\frac{1}{2}\] from this you can find \(\theta\)

Answer 7

theta is the arc tan \[\frac{ \sqrt{3} }{ 3 }\]

Answer 8

right?

Answer 9

so how do i put this into exponential form from there?

Answer 10

careful here!!

Answer 11

use sine and cosine

Answer 12

find the point on the unit circle with the coordinate
\[(\frac{\sqrt3}{2},-\frac{1}{2})\]

Answer 13

yes \[\frac{ 11\pi }{ 6}\]

Answer 14

but whre does the exponent come in?

Answer 15

are you trying to write this as
\[\large \sqrt3-i=re^{i\theta}\]?

Answer 16

or as
\[\sqrt3-i=r(\cos(\theta)+i\sin(\theta))\]?

Answer 17

the first one using euler's formula

Answer 18

ok well it is the same exactly

Answer 19

the \(r\) is still \(r\) and same \(\theta\)

Answer 20

you have a choice of angles

Answer 21

but if you pick \(\frac{11\pi}{6}\) then it is
\[\large 2e^{\frac{11\pi}{6}i}\]

Answer 22

ooooohhhhh i didn't know it was this simple! thank you

Answer 23

lol simple when you know how right? like most things...

Answer 24

yw