# Thread: Imaginary numbers with trig

1. ## Imaginary numbers with trig

Hi, I wasn't too sure what category to put this in, so i'm sorry if pre-calculus is wrong.

I came across this question but can't seem to prove it:

If $\displaystyle z=Cos\theta + \iota Sin\theta$, show that:

(i) $\displaystyle |1+z|=2Cos\frac{\theta}{2}$, for $\displaystyle 0^{\circ} \leq \theta \leq 180^{\circ}$

(ii) $\displaystyle \frac{2}{1+z}=1-\iota tan\frac{\theta}{2}$, $\displaystyle z\neq -1$

My workings:
(i)
$\displaystyle |1+z|$

$\displaystyle Cos0+\iota Sin0+Cos\theta +\iota Sin\theta$

$\displaystyle 2Cos\frac{\theta}{2}Cos\frac{-\theta}{2}+2\iota Sin\frac{\theta}{2}Cos\frac{\theta}{2}$

$\displaystyle Cos^2\frac{\theta}{2}+\iota Sin\frac{\theta}{2}Cos\frac{\theta}{2}=Cos\frac{\t heta}{2}$

After this I couldn't find a way to prove it and I just kept going round in circles.

(ii)
$\displaystyle \frac{2}{1+z}$

$\displaystyle \frac{1}{Cos\frac{\theta}{2}} = 1-\iota tan\frac{\theta}{2}$ ... from part one

$\displaystyle Sec^2\frac{\theta}{2} = 1-tan^2\frac{\theta}{2}-2\iota tan\frac{\theta}{2}$

$\displaystyle Sec^2A = 1-(Sec^2A-1)-2\iota tanA$

$\displaystyle Sec^2A = 1-\iota tanA$

$\displaystyle tan^2A = -\iota tanA$

$\displaystyle tanA = -\iota$

Which I don't think's possible.
Any help here would be appreciated.
Thanks.

2. Originally Posted by FreeT
Hi, I wasn't too sure what category to put this in, so i'm sorry if pre-calculus is wrong.

I came across this question but can't seem to prove it:

If $\displaystyle z=Cos\theta + \iota Sin\theta$, show that:

(i) $\displaystyle |1+z|=2Cos\frac{\theta}{2}$, for $\displaystyle 0^{\circ} \leq \theta \leq 180^{\circ}$

(ii) $\displaystyle \frac{2}{1+z}=1-\iota tan\frac{\theta}{2}$, $\displaystyle z\neq -1$

My workings:
(i)
$\displaystyle |1+z|$ Don't put the absolute value signs around the 1+z. At this point, you are finding an expression for 1+z, not |1+z|.

$\displaystyle Cos0+\iota Sin0+Cos\theta +\iota Sin\theta$

$\displaystyle 2Cos\frac{\theta}{2}Cos\frac{-\theta}{2}+2\iota Sin\frac{\theta}{2}Cos\frac{\theta}{2}$
That's correct up to that point. Also, $\displaystyle \cos(-x) = \cos x$, so you can write that last line as $\displaystyle 1+z = 2\cos^2\tfrac\theta2 + 2i\sin\tfrac\theta2\cos\tfrac\theta2 = 2\cos\tfrac\theta2(\cos\tfrac\theta2 + i\sin\tfrac\theta2)$.

Now remember that the absolute value of a complex number is given by $\displaystyle |x+iy| = \sqrt{x^2+y^2}$. Therefore $\displaystyle |1+z| = \sqrt{(2\cos\tfrac\theta2)^2(\cos^2\tfrac\theta2 + \sin^2\tfrac\theta2)}$.

3. Thanks. I got it out.

4. Originally Posted by FreeT
(ii) $\displaystyle \frac{2}{1+z}=1-\iota tan\frac{\theta}{2}$, $\displaystyle z\neq -1$

(ii)
$\displaystyle \frac{2}{1+z}$

$\displaystyle \frac{1}{Cos\frac{\theta}{2}} = 1-\iota tan\frac{\theta}{2}$ ... from part one

$\displaystyle Sec^2\frac{\theta}{2} = 1-tan^2\frac{\theta}{2}-2\iota tan\frac{\theta}{2}$

$\displaystyle Sec^2A = 1-(Sec^2A-1)-2\iota tanA$

$\displaystyle Sec^2A = 1-\iota tanA$

$\displaystyle tan^2A = -\iota tanA$

$\displaystyle tanA = -\iota$

Which I don't think's possible.
Any help here would be appreciated.
Thanks.
$\displaystyle \frac{2}{1 + z} = \frac{2}{1 + \cos{\theta} + i\sin{\theta}}$

$\displaystyle = \frac{2(1 + \cos{\theta} - i\sin{\theta})}{(1 + \cos{\theta} + i\sin{\theta})(1 + \cos{\theta} - i\sin{\theta})}$

$\displaystyle = \frac{2(1 + \cos{\theta} - i\sin{\theta})}{(1 + \cos{\theta})^2 + \sin^2{\theta}}$

$\displaystyle = \frac{2 + 2\cos{\theta} - 2i\sin{\theta}}{1 + 2\cos{\theta} + \cos^2{\theta} + \sin^2{\theta}}$

$\displaystyle = \frac{2 + 2\cos{\theta} - 2i\sin{\theta}}{1 + 2\cos{\theta} + 1}$

$\displaystyle = \frac{2 + 2\cos{\theta} - 2i\sin{\theta}}{2 + 2\cos{\theta}}$

$\displaystyle = 1 - \frac{2i\sin{\theta}}{2 + 2\cos{\theta}}$

$\displaystyle = 1 - \frac{i\sin{\theta}}{1 + \cos{\theta}}$

$\displaystyle = 1 - i\tan{\frac{\theta}{2}}$.