Complex Solution to an Exponential Equation

[cdummie]

Homework Statement

Solve the following equation: $(1+a)^n=(1-a)^n$ where a is complex number and n is natural number

Homework Equations

Euler's formula

The Attempt at a Solution

I've tried something like this

$(1+a)^n=(1-a)^n \\ (\frac{1+a}{1-a})^n=1$

But i really have no idea what to do next, since that n-th degree complicates things, i mean i could write a as a= x+iy where x and y are real numbers but it won't help much.

 

[mfb]

What do you know about numbers z that satisfy zⁿ = 1? In particular, how many are there?

 

[cdummie]

What do you know about numbers z that satisfy zⁿ = 1? In particular, how many are there?

Well, there are n solutions to such equation, $z=cos\frac{2\pi +2k\pi}{n} + isin\frac{2\pi +2k\pi}{n}$

 

[mfb]

Right. Using the $r e^{i \phi}$ notation is probably easier in the next steps.

 

[cdummie]

Right. Using the $r e^{i \phi}$ notation is probably easier in the next steps.

Ok, so, i have $\frac{1+a}{1-a}=e^{i\frac{2k\pi}{n}}$ , but i should find a, what should i do next? Since a is complex number a=x +iy (x, y are real numbers) and then if i multiply and divide LHS by by 1-x +iy it doesn't seem like it gets me any closer to the solution.

 

[HallsofIvy]

Why not? $e^{i\frac{2k\pi}{n}}$ is a constant. If it is bothering you, call it "k". Then $\frac{1+ a}{1- a}= k$ which we can write 1+ a= k(1- a)= k- ka. Can you solve that equation for a? Then, if you need to, write a as x+ iy to solve for x and y.

 

[cdummie]

Why not? $e^{i\frac{2k\pi}{n}}$ is a constant. If it is bothering you, call it "k". Then $\frac{1+ a}{1- a}= k$ which we can write 1+ a= k(1- a)= k- ka. Can you solve that equation for a? Then, if you need to, write a as x+ iy to solve for x and y.

Well, i think it could go this way:

$1+a=k-ka \\ a+ka=k-1 \\ a=\frac{k-1}{k+1} \\ a=\frac{e^{i\frac{2k\pi}{n}}-1}{e^{i\frac{2k\pi}{n}}+1}$

Is this correct?

 

[Samy_A]

Well, i think it could go this way:

$1+a=k-ka \\ a+ka=k-1 \\ a=\frac{k-1}{k+1} \\ a=\frac{e^{i\frac{2k\pi}{n}}-1}{e^{i\frac{2k\pi}{n}}+1}$

Is this correct?

Looks on track, but ...
What happens when $\frac{k}{n}=\frac{1}{2}$?

 

[cdummie]

Looks on track, but ...
What happens when $\frac{k}{n}=\frac{1}{2}$?

I think i know what are you trying to say, denominator can't be zero, so this works for every k and n except when n=2k.

 

[PeroK]

I think i know what are you trying to say, denominator can't be zero, so this works for every k and n except when n=2k.

Yes, but can you simply that expression you have?

 

[Samy_A]

I think i know what are you trying to say, denominator can't be zero, so this works for every k and n except when n=2k.

Indeed, because in your derivation you at a point divided by $e^{i\frac{2k\pi}{n}}+1$, and that doesn't work if that value is 0.
As @PeroK suggested, you can simplify your result.

 

[cdummie]

Yes, but can you simply that expression you have?

Let's see, i guess i could multiply numerator and denominator by $cos\frac{2k\pi}{n} + 1 - sin\frac{2k\pi}{n}$ is this the thing i should do?

 

[mfb]

There is no need to start with cosine and sine. How can you expand every fraction to make its denominator real?

 

[cdummie]

There is no need to start with cosine and sine. How can you expand every fraction to make its denominator real?

Well, if it's complex number, i can do that by multiplying with it's conjugate, that's what i did, i can't remember something else.

 

[mfb]

Right, multiply both sides with its complex conjugate (but don't use sine and cosine, they just make it messy).