Homework Help: Couple of complex number questions

1. Sep 17, 2008

rock.freak667

1. The problem statement, all variables and given/known data

1)If either |z|=1 or |w|=1,prove that

$$|\frac{z-w}{1-\overline{z}w}|=1$$

2)If z1,2,3, are complex numbers and

$$\frac{z_2 - z_1}{z_3 -z_1}=\frac{z_1 -z_3}{z_2-z_3}$$

Show that |z2-z1|=|z3-z1|=|z2-z_3|

3) Find the sum of the following series:

nC1sinx + nC2 sin2x + nC3 sin3x +...+ nCn sin(nx)

2. Relevant equations

$$|\frac{Z_2}{Z_1}|=\frac{|Z_2|}{|Z_1|}$$

3. The attempt at a solution

(Will type in the next post since, the site is giving trouble for me at the moment)

2. Sep 17, 2008

rock.freak667

For the questions, I didn't reach too far but I will put what I did.

1)$$|\frac{z-w}{1-\overline{z}w}|$$

$$\Rightarrow \frac{|z-w|}{|1- \overline{z}w|}$$

Don't know how to get past this, only thing I can add in is that |z|=|$\overline{z}$|

2)$$\frac{z_2 - z_1}{z_3 -z_1}=\frac{z_1 -z_3}{z_2-z_3}$$

So far all I've reached is

$$(z_2-z_1)(z_2-z_3)=-(z_1-z_3)^2$$

3)nC1sinx + nC2 sin2x + nC3 sin3x +...+ nCn sin(nx)

I don't know how to deal with the nC1,nC2,etc. If they binomial coefficent terms weren't there I could easily do it but the nC1,nC2, are confusing me on how to approach the problem.

3. Sep 17, 2008

Dick

For the first one, if you multiply z and w by the same complex phase (i.e. rotate by the same angle) the absolute value doesn't change. The denominator remains the same and the numerator rotates by that phase. So if |z|=1 you can rotate it so that z=1. Ditto for w. It easy to figure out the z=1 or w=1 cases.

4. Sep 17, 2008

rock.freak667

I don't understand the rotation thing. I get that no matter what the argument, the modulus will be the same. But I don't get how to multiply z and w by the same complex phase.

5. Sep 17, 2008

Dick

I mean if you replace z -> e^(it)*z, w -> e^(it)*w, the absolute value doesn't change. So for example if |z|=1, pick t so that e^(it)*z=1.

6. Sep 18, 2008

rock.freak667

Well for e^(it)*z=1, t must be 0,2pi,4pi,etc.

So then it's just

$$\frac{|ze^{i2 \pi}-w|}{|1-e^{i2 \pi} \overline{z}w|}$$

7. Sep 18, 2008

Dick

No it doesn't. It becomes |z-w|/|1-(z*)w|=|e^it*(z-w)|/|1-(z*)e^(-it)*e^(it)*w|=|1-e^(it)w|/|1-1*e^(it)w|.

8. Sep 18, 2008

rock.freak667

So we have to replace z by ze^(it) and w by we^(it)t.

But I don't see how you got this |1-e^(it)w|/|1-1*e^(it)w|. From this |e^it*(z-w)|/|1-(z*)e^(-it)*e^(it)*w|.

9. Sep 18, 2008

Dick

I don't know why you are fighting this. Try a different notation. If |z|=1 then z(z*)=1 and |z*|=1. 1*|z-w|/|(1-(z*)w)|=|z*||z-w|/|1-(z*)w)|=|(z*)(z-w)|/|1-(z*)w|=|1-(z*)w|/|1-(z*)w|=1!!!!!! Regardless of how you notate it the notion is still the same. The absolute value is invariant under a rotation of w and z by the same complex phase. Use that freedom to rotate the one whose absolute value is one to become 1.

10. Sep 18, 2008

rock.freak667

Now I get it...didn't understand the e^(it) thing. But this I get now. Thanks for this.

Any hints for the 2nd and 3rd one?

11. Sep 18, 2008

Dick

No, I've been too busy arguing over the first one. I'll give them some thought though.

12. Sep 18, 2008

Dick

I really hate to break this to you but the only reasonable way I can think of to solve 2) is a version of the solution to 1). Promise not to quarrel, ok? The problem is translation invariant. You can replace z1, z2 and z3 by z1-c, z2-c and z3-c for any c and it's still true. It's also scale invariant, you can replace z1, z2 and z3 by z1/a, z2/a and z3/a for any a and it's still true. You can use these invariances to replace, say, z1=0 and z2=1. Now you can actually solve for the possibilities for z3 and check that |z1-z2|=|z2-z3|=|z1-z3|. I.e. z1, z2 and z3 lie on an equilateral triangle. It's possible that I'm just thinking like a physicist here and there is a more elegant solution, but I'm not seeing it. Let me know if you find one.

13. Sep 18, 2008

Dick

Where are you getting these problems? That's actually kind of 'subtle'.

14. Sep 18, 2008

rock.freak667

Coursework from my engineering class.

Well I guess I'll try not to over confuse myself with the 2nd one.

15. Sep 18, 2008

Dick

I wouldn't give that to an engineering class. Think of it this way. First move the origin to z1. Now scale by dividing everything by the resulting z2. So you can solve the new problem by taking z1=0 and z2=1 and finding z3.

16. Sep 19, 2008

Dick

For the third one, that looks like the imaginary part of (1+exp(i*x))^n, doesn't it? In fact, it is! So now you just have to find a nice way of writing (1+exp(i*x))^n. Try putting (1+exp(i*x)) in polar form. Thanks to some trig identities, you can. The most obscure one is sin(x)/(1+cos(x))=tan(x/2).

17. Sep 19, 2008

rock.freak667

Can I ask how you came up with this? Well the nCr coefficents would indicate a binomial but why 1+exp(ix)? Could I have just used any constant+exp(ix)?

18. Sep 19, 2008

Dick

(1+a)^n=sum(nCk*1^(n-k)*a^k, k=0 to n). If a=exp(i*x) then a^k=exp(i*k*n). The imaginary part of exp(i*k*n)=sin(k*n). If you had used any constant, then you would be getting powers of that constant.

19. Sep 20, 2008

rock.freak667

$$|1+e^{ix}|=\sqrt{2(1+cosx)}$$

$$arg(1+e^{ix})=\frac{x}{2}$$

so then the required sum is just: Im{1+eix} which is:

$${2(1+cosx)}^\frac{n}{2} sin\frac{nx}{2}$$

Last edited: Sep 20, 2008
20. Sep 20, 2008

Dick

Seems about right to me. Seem ok to you?

21. Sep 21, 2008

rock.freak667

yeah it is ok.

For the second one, I think I am supposed to say that the arg(LHS)=arg(RHS) and therefore it is an isosceles triangle and then from that I need to draw the complex representation of one of the quotients and then try to prove that it's an equilateral triangle.

22. Sep 21, 2008

Dick

That will work. So what do you conclude from isosceles? You also have more information from |LHS|=|RHS|. Use that too.

23. Sep 21, 2008

rock.freak667

Here is something else I tried, I don't know if it is correct though.

$$arg(\frac{z_2-z_1}{z_3-z_1})=arg(\frac{z_1-z_3}{z_2-z_3})$$

$$arg(z_2-z_1)-arg(z_3-z_1)=arg(z_1-z_3)-arg(z_2-z_3)$$

$$arg(z_2-z_1)-arg(z_3-z_1)= \pi + arg(z_3-z_1)-arg(z_2-z_3)$$

$$arg(z_2-z_1)-2arg(z_3-z_1)+arg(z_2-z_3)= \pi$$

$$arg(\frac{(z_2-z_1)(z_2-z_3)}{(z_3-z_1)^2})= \pi$$

$$\frac{(z_2-z_1)(z_2-z_3)}{(z_3-z_1)^2}= tan(\pi)=0$$

$$\Rightarrow (z_2-z_1)(z_2-z_3)=0$$

and so $z_2=z_1$,$z_2=z_3$

Hence $z_1=z_2=z_3$

Can I just subract appropriately and then say that all the modulii are equal? For example, if I subtract z1 I'll get

$0=z_2-z_1=z_3-z_1 \Rightarrow |z_2-z_1|=|z_3-z_1|$