Complex Numbers - Complex Roots of Unity

[shabi]

Need help with this please:

Homework Statement

(1 + cosθ + isinθ) / (1 - cosθ - isinθ) = icotθ/2

The first step in the solutions shows:

(2cos^2θ/2 + i2sinθ/2cosθ/2) / (2sin^2θ/2 - i2sinθ/2cosθ/2)

Homework Equations

I can't get there.

The Attempt at a Solution

I tried multiplying by: (1 - cosθ + isinθ) / (1 - cosθ + isinθ), with no luck.

the top line of my attempt shows i2sinθ + i2sinθcosθ
because 1-cos^2θ=sin^2θmy signs could be wrong but still. how does the θ become θ/2. this is doing my head in. maybe just sleep on it.

 

[Curious3141]

Need help with this please:

Homework Statement

(1 + cosθ + isinθ) / (1 - cosθ - isinθ) = icotθ/2

The first step in the solutions shows:

(2cos^2θ/2 + i2sinθ/2cosθ/2) / (2sin^2θ/2 - i2sinθ/2cosθ/2)

Homework Equations

I can't get there.

The Attempt at a Solution

I tried multiplying by: (1 - cosθ + isinθ) / (1 - cosθ + isinθ), with no luck.

the top line of my attempt shows i2sinθ + i2sinθcosθ
because 1-cos^2θ=sin^2θmy signs could be wrong but still. how does the θ become θ/2. this is doing my head in. maybe just sleep on it.

OK, you know from the standard double-angle formulae that $\sin{2\alpha} = 2\sin\alpha\cos\alpha$ and that $\cos{2\alpha} = 2\cos^2{\alpha} - 1 = 1 - 2\sin^2{\alpha}$.

Now put $\alpha = \frac{\theta}{2}$.

The results you get are often called the half-angle formulae, but it's not worth remembering them specifically because they're so easily derived from the double-angle formulae.

 

[shabi]

Thanks for the prompt reply and pointing that out!

So simple now i can see that.

 

[Curious3141]

Thanks for the prompt reply and pointing that out!

So simple now i can see that.

You're welcome.