Question

Find the two square roots for the complex number. Answer in standard form.
-2+2i√3

Answer 1

@Ala123 Can you convert this into polar form?

Answer 2

you mean Trig form ?

Answer 3

The form
\[re^{i \theta}\]

Answer 4

what form is that? i only learned trig form for this

Answer 5

Ok, write that down

Answer 6

Yeah, just remembered you can use that also

Answer 7

idk how to do that :|

Answer 8

is that re^itheta ?

Answer 9

You write that trigonometric form. We'll work with that

Answer 10

(a+i*b)²=a²-b² +i*2*a*b=-2+2i√3 compare real and imaginary, recall a,b element from R

Answer 11

ok 1 step at a time please -.-

Answer 12

i know if you convert to trig form you gotta find r first

Answer 13

r=√a^2+b^2

Answer 14

in this case a^2 =-2 but what is b^2 ?

Answer 15

b=2√3i ?

Answer 16

I'll help you with that

Answer 17

a=-2 and \(b=2\sqrt 3\)
Don't include i

Answer 18

@ash2326 if you put the 2√3 and you square it mean the root cancel out right ?

Answer 19

Yeah
\[ (2 \sqrt 3)^2= 2 \times 2 \times \sqrt 3\times \sqrt 3\]

Answer 20

is it 6 ?

Answer 21

oh it's 7

Answer 22

Is this r?

Answer 23

yes

Answer 24

oh no i mean r=root 7

Answer 25

root 11

Answer 26

r=root 11

Answer 27

How did you get that?
\[r=\sqrt {2^2+ (2\sqrt 3)^2}=\sqrt {4+12}=4\]

Answer 28

r=4

Answer 29

there lol lol =.=

Answer 30

ok, find the theta?

Answer 31

theta can you just put tan inverse(b/a)

Answer 32

i got theta=60 :D

Answer 33

It's better to plot this first. If you plot this point then you'll know better what's theta?

Answer 34

120 ?

Answer 35

Yay:D that's right, now what's the r e^(itheta form)

Answer 36

i only know Trig form =.= which is 4(cos120+ i sin120)

Answer 37

I made a mistake sorry:(
Theta is not 120
Thanks @satellite73 for pointing out

Answer 38

huh?

Answer 39

@ash2326 no you are right, \(\theta =120\) if you are working in degrees.

Answer 40

yeah, lol u just gave me a mini heart attack :O

Answer 41

take the square root of 4, take half of 120
get
\[2(\cos(60)+i\sin(60))\] evaluate

Answer 42

then go half way across the circle

Answer 43

2cis60 and 2cis240 and idk how to turn into standard form :D

Answer 44

\[2(\cos60+isin60)=2(\frac{1}{2}+i\frac{\sqrt{3}}{2})=1+\sqrt{3}i\]

Answer 45

oh yeah :D

Answer 46

\[2(\cos 240+i \sin 240)=2(\frac{-1}{2}-\frac{\sqrt{3}}{2})=-1-\sqrt{3}\]

Answer 47

i. I forgot the i

Answer 48

ok i'll include it thanks Mertsj

Answer 49

yw