Question

how can i make -1+sqrt(3i) into trig form

Answer 1

@wio let's solve this!!

Answer 2

Never tried these types of problems before, I'm interested. I don't know where to begin though, lol.

Answer 3

It is\[
-1+\sqrt 3 i
\]Or is it: \[
-1+\sqrt{3i}
\]

Answer 4

I believe the \(i\) should not be in the square root, but if it is we can still do this.

Answer 5

Shall we try it both ways?

Answer 6

Okay, we can do that.

Answer 7

So the trig form would is like: \[
a+bi = re^{i\theta} = r[\cos(\theta) + i\sin(\theta)]
\]

Answer 8

http://openstudy.com/study#/updates/530bfd53e4b03fec1e318cfc was looking at previous examples.

Answer 9

Basically, \((a,b)\) represent the Cartesian coordinates, while the polar coordinates are \((r,\theta)\).

Answer 10

Oh i see.

Answer 11

So: \[
r = \sqrt{a^2+b^2} \\
\theta =\tan^{-1}\left(\frac ab\right)
\]

Answer 12

It really is that simple.

Answer 13

I thought it'd be a little bit more complex...haha.

Answer 14

One thing to notice is that \(a^2+b^2 = z\bar z\) where \(z=a+bi\) and \(\bar z=a-bi\) (called the conjugate).

Answer 15

We call \(r=\sqrt{z\bar z}\) the modulus or \(z\), for some reason, and we call \(\theta\) the argument.

Answer 16

Oh okay. Makes sense.

Answer 17

so \[
r = \sqrt{1^2+\sqrt{3}^2} = \sqrt{1+3} = 2
\]

Answer 18

Whoops, I messed up earlier. I should have said: \[
\theta = \tan^{-1}\left(\frac ba\right)
\]

Answer 19

b/a?... Oh yeah, similar to y/x

Answer 20

In fact, it would be easier to just say \[
z=x+yi
\]and use \((x,y)\). By convention \((a,b)\) is more popular.

Answer 21

I don't like complex analysis, because we should just be learning about 2D vectors instead.

Answer 22

Hmm.

Answer 23

The only thing about complex analysis that is interesting is complex multiplication. There isn't any vector equivalent.

Answer 24

Well, maybe a couple other things as well. I never fully did complex analysis.

Answer 25

I didn't know that's what this was xD

Answer 26

Anyway \[
\theta = \tan^{-1}\left(\frac{\sqrt3}{-1}\right)=-\frac{\pi}{3} \left(=-60^\circ\right)
\]

Answer 27

\[
-1+\sqrt 3i = 2\left[\cos\left(-\frac\pi 3\right)+i\sin\left(-\frac\pi 3\right)\right]
\]

Answer 28

Wait, why is \(a=1\)? Wouldn't \(a=1\)?

Answer 29

\[
r = \sqrt{(-1)^2+(\sqrt{3})^2} = \sqrt{1+3} = 2
\]

Answer 30

Oh that's right. Okay.

Answer 31

In the case of: \[
-1+\sqrt{3i}
\]You basically have to find the \(\sqrt{i}\), which actually has two solutions.

Answer 32

Heyoooo, Euler's Formula, I *do* recognize at least some of this, lol.

Answer 33

\[
i = (x+yi)^2=(x^2-y^2)+(2xy)i
\]

Answer 34

This means\[
x^2-y^2=0,\quad 2xy = 1
\]

Answer 35

since we can say \(i = (0) + (1)i\)

Answer 36

I'm little stuck! not understanding where the 0 and 1 are coming from...>.<

Answer 37

Well, basically \(i\) is a complex number of the form \(x+yi\) where \((x,y) = (0,1)\).

Answer 38

Oh okay. I thought you derived it from something else. Nvm!

Answer 39

\[
x^2-y^2=0\implies x^2=y^2 \implies |x|=|y| \implies x=y,x=-y
\]Plug it into the other equation an: \[
2yy=1,2(-y)y = 1
\]

Answer 40

Oh cool

Answer 41

We can't let \(x\) or \(y\) be imaginary numbers, so the only equation that matters is: \[
2y^2=1 \implies y=\pm \sqrt{\frac{1}{2}}
\]

Answer 42

Why can we not let them be imaginary?

Answer 43

Because they come from: \[
x+yi\], which is already a complex number, with real coefficients.

Answer 44

Ohh. I see that connection.

Answer 45

Okay, so let's test it out real quick: \[
\left(\frac 1{\sqrt 2} + \frac 1{\sqrt{2}}i \right)^2
\]

Answer 46

\[
=\frac 12 +\frac 12i+\frac 12i -\frac 12 = \left(\frac 12 -\frac 12\right) +\left(\frac 12+\frac 12\right)i = 0+1i = i
\]

Answer 47

If that is a solution, then so is: \[
-\frac 1{\sqrt2}-\frac{1}{\sqrt 2}i
\]

Answer 48

nvm that would result in something completely different.

Answer 49

No, if \(z^2=i\) then \((-z)^2=i\). This works for complex numbers as well.

Answer 50

Ahh, interesting. O_o

Answer 51

We ended up with\[
x=y\\
2y^2=1
\]So we know \(x\) and \(y\) must have the same sign and absolute value.

Answer 52

I see. Got it.

Answer 53

Another way to do this, with less algebra, is to use trig coordinates.

Answer 54

\[
i=(0)+i(1)=1[\cos(\pi/2)+i\sin(\pi/2)]=1e^{i\pi/2} = e^{i\pi/2}
\]

Answer 55

Take the square root, and we get: \[
\pm\sqrt{e^{i\pi/2}} = \pm e^{i\pi/4} = \pm[\cos(\pi/4)+i\sin(\pi/4)] = \pm\left[\frac 1{\sqrt2}+\frac 1{\sqrt2}i\right]
\]

Answer 56

The only reason all of this works is due to the fact that: \[
[\cos(\theta)+i\sin(\theta)]^n = \cos(n\theta)+i\sin(n\theta)
\]

Answer 57

And this formula works due to than double angle formula: \[
\cos(2\theta) = \cos^2(\theta)-\sin^2(\theta)\\
\sin(2\theta) = 2\cos(\theta)\sin(\theta)
\]

Answer 58

Ah, I don't think i'm comfortable with the function of \(i\) as of yet. I have to keep referencing back to your post describing what it is. \[i = (x+yi)^2=(x^2-y^2)+(2xy)i \iff x^2-y^2=0,\quad 2xy = 1\]

Answer 59

I understand how it equates to 0 and 1, but now i'm confused with how you got to \[=1[\cos(\pi/2)+i\sin(\pi/2)]=1e^{i\pi/2} = e^{i\pi/2}\] That part.

Answer 60

Well, I wanted:\[
(0)+i(1) = \cos(\theta) + i\sin(\theta)
\]where \[
\cos(\theta) =0,\quad \sin(\theta) =1
\]

Answer 61

The only places where \(\cos(\theta)=0\) (in the first revolution \(\theta \in [0,2\pi)\)), is \(\theta=\pi/2\) and \(\theta=3\pi/2\).

Answer 62

Ok, i get that.

Answer 63

That gives \(\sin(\pi/2)=1\) and \(\sin(3\pi/2)=-1\)... so we say \(\theta=\pi/2\).

Answer 64

Then \[
(0)+i(1) = \cos(\pi/2)+i\sin(\pi/2).
\]

Answer 65

Understood.

Answer 66

I basically just converted \(i\) to its trig coordinates. Since you have sort of learned it a moment ago, I figured you would get it. But you're not completely familiar yet so I was going too fast.

Answer 67

Yeah a little bit.

Answer 68

But after that explanation I found the connection.

Answer 69

Actually, for \(i\) my method for finding \(\theta\) breaks down a bit since you are using: \[
\theta = \tan^{-1}\left(\frac 10\right)
\]

Answer 70

I see.

Answer 71

You just have to realize that when \[
\tan(\theta) = \pm \infty
\]Then it gives you \[
\theta = \pm \pi/2 + 2\pi n
\]

Answer 72

Well yeah. I just had to visualize it.