# Complex Numbers

1. Oct 9, 2007

### Howers

1) A mathemetician is willing to sell you something valued at \$i^i. Would you pay him 20 cents for it?

2) Let z=(z1/z2) where z1 = a+ib and z2 = c+id. Show the angle of z is the difference between angle z1 and z2.

3) Show that multiplying any vector by e^ix doesn't alter its length.

My attempts:

1) Tried using cosx + isinx to no avail. Honestly have no idea where to begin.
2) First I found the z by multiplying it out. let x be the angle. so i did tanx = bc-ad/ac+bd. Individually, you get tanx1 = b/a and tanx2=d/c which is not the same.
3) in the length formula i end up with cos^2-sin^2 which does alter length.

ruined my thanks giving, so i give up onthese.

2. Oct 9, 2007

### neutrino

You're more or less on the right track...use Euler's formula/identity.

Again, this is correct. You just need to show that difference of the angles is equal to arctan((bc-ad)/(ac+bd)). Look up trig. identites.

I don't quite understand what question 3 asks for. Sorry.

3. Oct 13, 2007

### atqamar

Euler's Identity

1)
So you're solving for $$i^i$$. The Euler's Identity states that $$e^{\pi i}+1=0$$. Using just this identity, many other quantities can be derived, including our $$i^i$$.

Start by subtracting $$1$$ from both sides, then getting the square root of both sides:
$$e^{\frac{1}{2}\pi i}=(-1)^{\frac{1}{2}}$$

$$(-1)^{\frac{1}{2}}$$ obviously can be written as $$\sqrt{-1}$$, which simplifies to $$i$$:
$$e^{\frac{1}{2}\pi i}=i$$

Now raise both sides to the power of $$\frac{1}{i}$$:
$$e^{\frac{1}{2}\pi i\frac{1}{i}}=i^{\frac{1}{i}}$$
$$e^{\frac{1}{2}\pi}=i^{\frac{1}{i}}$$

Since $$\frac{1}{i}$$ equals $$i^{-1}$$, substitute:
$$e^{\frac{1}{2}\pi}=i^{i^{-i}}$$

Now simply raise both sides to the power of $$-1$$:
$$e^{-\frac{1}{2}\pi}=i^{i}$$

And voila: $$i^{i} = e^{-\frac{1}{2}\pi} = 0.207879576 ...$$

Now you be the economist and tell me whether you will accept the offer or not.

4. Oct 13, 2007

### huckmank

Close. We want to have the magnitude (absolute value) of e^ix = 1. So we write e^ix in the form of cos(x)+isin(x). Finding the magnitude of this is done by finding the square root of the sum of the squares so sqrt(cos(x)^2 + sin(x)^2). You'll notice that the i dropped out. This is because the definition of the magnitude of a set of terms is simply sqrt(t1^2+t2^2+...tn^2). We don't care that i^2 is -1, we still sum the term.

I think you'll agree that cos(x)^2+sin(x)^2 is 1 and that sqrt(1) is 1.

To the mods, the only reason I gave the full answer was that the OP was basically there, s/he just needed to see why it was positive sin(x)^2.

5. Oct 14, 2007

### Gib Z

For 3, the simplest way to do it is to write the general complex number in polar form:

$$z= r e^{i\theta}$$.

Which quantity gives the length? Does multiplying by $e^{ix}$ affect this quantity?

6. Oct 14, 2007

### real10

for second question a quicker way is to convert to the polar form and then u can see the angle of z1/z2

for first question i=e^i*pi/2 (since sine(pi/2)=1) raise this to "i" power u immediately get e^-pi/2.

Last edited: Oct 14, 2007
7. Oct 14, 2007

### Gib Z

Good work, but try to stick to one notation for the imaginary unit at a time. Are you a physics or engineering student?

8. Oct 14, 2007

### real10

EDIT: everything changed to i now

9. Oct 14, 2007

### huckmank

That's not really a proof. That's saying that e^xi doesn't affect the length because... well it doesn't affect the length.

10. Oct 14, 2007

### Gokul43201

Staff Emeritus
I disagree. You can always calculate the norm of the vector before and after, and show that it is the same.