Question

Use Euler’s formula to rewrite this radical expression in exponential form.

2√2 - 2i√2

Answer 1

I have Euler's Formula as:\[a+bi=r(\cos \Theta+i \sin \Theta)\]

But how do i plug it in...??

Answer 2

This is not Euler's Formula?

Answer 3

I don't know much about this; but do you have formulas for the values of r and θ?

Answer 4

My textbook says that's the formula for a complex number, a+bi.

I also have \[e^(i \alpha) = \cos \alpha+i \sin \alpha \]

but that makes even less sense.

This is literally all the information i am given. that's why i'm confused.

Answer 5

Oh I understand now; you have to express \(2\sqrt2-2i\sqrt2\) in the form of \(r(\cos\theta+i\sin\theta)\) first.

Answer 6

.... how do i do that? :(

Answer 7

To express \(a+bi\) in the form of \(r(\cos\theta+i\sin\theta)\):
\[r=\sqrt{a^2+b^2}\]\[\tan\theta=\frac ba\]

Answer 8

Or you can solve equations:
\[a=r\cos\theta\]\[bi=ri\sin\theta\]

Answer 9

And you'll come up with my formulas

Answer 10

Go with Kenny's method, it's more reliable,
but here's another idea! This is what I like to do at least :)
\[\Large\rm \color{royalblue}{2\sqrt2} - \color{orangered}{2\sqrt2}\mathcal i\]
Try to turn each term into the special values that correspond to the unit circle.
Factoring a 4 out of each term,\[\Large\rm 4\left(\color{royalblue}{\frac{\sqrt2}{2}}-\color{orangered}{\frac{\sqrt2}{2}}\mathcal i\right)\]
Then relating it to Euler,\[\Large\rm \color{royalblue}{\cos \theta=\frac{\sqrt2}{2}},\qquad\qquad\qquad \color{orangered}{\sin \theta=-\frac{\sqrt2}{2}}\]Sine is negative, cosine is positive, uhhh we must be in the 4th quadrant I guess.
And we sqrt2's, so we must be dealing with one of the pi/4's right?

Answer 11

Okay so i need to get this in exponential form...
So, after i get it to \[4(\cos 7\Pi/4 + i \sin 7\Pi/4)\] how do i get it to exponential form???

Answer 12

\[\Large\cos\theta+i\sin\theta=e^{i\theta}\]

Answer 13

... i dont understand x(

Answer 14

\(\theta=\dfrac{7\pi}4\)

Answer 15

\[\Large\cos\frac{7\pi}4+i\sin\frac{7\pi}4=e^{i\frac{7\pi}4}\]

Answer 16

OOOHHH, I SEE! Hahaha wow, okay, yeah. Thank you soooo so much! :))))

Answer 17

no problem :)