Question

Here is my second question on the complex numbers assignment. How do I prove the following below?

If z=cos`theta`+isin`theta`, prove that;
1. 1+`z+z^2`=(1+`cos theta)(cos theta+isin theta)`
2. `z//(1+z)=1+itan(theta//2)`

Answer 1

See below

Explanation:

1. `z= costheta+isintheta`

`z+1= costheta+isintheta+1`

`(z+1)^2=(costheta+isintheta+1)^2`

`1+2z+z^2= cos^2theta+2costhetaisintheta+2costheta+i^2sin^2theta+2isintheta+1`

`1+2z+z^2= cos^2theta+2costhetaisintheta+2costheta-sin^2theta+2isintheta+sin^2theta+cos^2theta`

`1+2z+z^2= 2cos^2theta+2costhetaisintheta+2costheta+2isintheta`

Factor:
`1+2z+z^2= 2(1+costheta)(costheta+isintheta)`

Answer 2

You asked this:

For 1), you are proving that:

• `1 + z + z^2 = z(1 + cos theta) qquad equiv qquad 1/z + 1 + z = 1 + cos theta`

`z` is the unit circle so I don't see a problem with dividing like that.

Well:

`qquad 1/z = barz/(z bar z) = (cos theta - i sin theta)/(cos^2 theta + sin^2 theta) = cos theta - i sin theta`

So:

`qquad 1/z + 1 + z = cos theta - i sin theta + 1 + cos theta + i sin theta`

`qquad = 2cos theta + 1`

`implies 1 + z + z^2 = (1 + bb2 cos theta)(cos theta + i sin theta)`

That's not what you're looking for but it is the same as other answer posted here for this question, if you actually finish off the algebra.

For 2) , I think the answer is out by a factor of 2:

`z/(1 + z) = (1 + z - 1)/(1+z)`

`= 1 - 1/(1+z)`

`= 1 - bar(1+z)/((1+z) bar((1+z)))`

`= 1 - (1 + cos theta - i sin theta )/((1 + cos theta)^2 + sin^2 theta)`

`= 1 - (1 + cos theta - i sin theta )/(2 + 2 cos theta )`

`= 1/2 + i( sin theta )/(2 + 2 cos theta )`

Half angle formulae:

`= 1/2 + i( 2 sin (theta/2) cos (theta/2) )/(2 + 2 (2 cos^2 (theta/2) - 1) )`

`= 1/2 + i( 2 sin (theta/2) cos (theta/2) )/( 4 cos^2 (theta/2) )`

`= bb(1/2)(1 + i tan (theta/2) )`

Again not the answer you're looking for but I don't see the mistake in the algebra.