Question

"Hard Complex Numbers" - Question
Medal!

Answer 1

The complex number w has modulus 1 and argument 2θ radians.
Show that
\[\frac{ w-1 }{ w+1 }=i \tanθ \]

Answer 2

oops the ������������������ is theta

Answer 3

Let x = theta...
I have tried plugging in w = cos2x + i sin 2x

Answer 4

I would plug in what you describe: \[\LARGE w=e^{i2 \theta}\] it will be easier to manipulate this way.

Answer 5

Ah ok, so we then get:
\[\frac{ e^{i2\theta}-1 }{ e^{i2\theta}+1 }\]
But how would you simplify that?
Sorry I'm really bad at simplifying...

Answer 6

Ok, multiply by 1 that looks like this then, this should make sense! \[\LARGE \frac{e^{-i \theta}}{e^{-i \theta}}\]

Answer 7

Huh what do you mean multiply by 1?

Answer 8

See that fraction? It's just the same thing divided by itself, just like 3/3=1 that is also 1.

Answer 9

Multiply by the equation you gave above? If so, how did you get it?

Answer 10

Yeap I get that but how did you know to multiply by that?

Answer 11

What I have in mind are the trig formulas for sine and cosine which look like this: \[\LARGE \frac{e^{i \theta}+e^{-i \theta}}{2}= \cos \theta\]

Afterall, we want this to look like tan theta right?

Answer 12

Oh yeah I understand that now

Answer 13

Ok cool, so now where do we stand? Have you figured it out now?

Answer 14

I'm trying it right now but I'm just stuck on what
\[e^{i2\theta} e^{-i \theta}\]
is equal to

Answer 15

@Kainui ?

Answer 16

This is just normal exponent rules. For example, if I give you \[\LARGE x^2*x^3=(x*x)*(x*x*x)=x^5\] the rule is just to add the exponents as we can see.

What is a negative exponent? Well suppose you had \[\LARGE \frac{x^3}{x^2}=x^1\] Since we already know exponents add, then it makes sense that we can rewrite this as \[\LARGE \frac{x^3}{x^2}=x^3*x^{-2}=x^1\] Similarly, this is why we also get the rule for x^0=1. \[\LARGE \frac{x^3}{x^3}=1=x^3*x^{-3}=x^{3-3}=x^0\]

Answer 17

So I get
[\frac{e^{i2\theta + i \theta} - e^{-i \theta}}\frac{ ? }{ ? }\]
\[\frac{ e^{i2\theta + i \theta} - e^{-i \theta} }{ e^{i2\theta + i \theta} + e^{-i \theta} }\]

Answer 18

What to do next?

Answer 19

\[\large \dfrac{e^{i2\theta}+1}{e^{i2\theta}-1}\times \dfrac{e^{-i\theta}}{e^{-i\theta}} =\dfrac{e^{i\theta}+e^{-i\theta}}{e^{i\theta}-e^{-i\theta}} \]

Answer 20

divide by 2 both top and bottom
then look at sinx, cosx definitions