# Thread: Help with Complex Number Question!

1. ## Help with Complex Number Question!

Not sure how to do this question:

If z = cos θ + isin θ, show that (1-z)/(1+z) = -itan(θ/2)

Any help would be appreciated. Thanks!

2. ## Re: Help with Complex Number Question!

1-z = 1 - cosx - isinx
1+z = 1+cosx+isinx

Times top and bottom by 1+cosx-isinx

Numerator = (1-cosx-isinx)(1+cosx-isinx)
= 1+cosx-isinx-cosx-cos^2x+isinxcosx-isinx-isinxcosx-sin^2x
= 1-2isinx-(sin^2x+cos^2x)
= -2isinx

Denominator = (1+cosx+isinx)(1+cosx-isinx)
= (1+cosx)^2 - (isinx)^2
= 1+cos^2x+2cosx+sin^2x
= 2(1+cosx)

(1-z)/(1+z) = -isinx/(1+cosx)
= -i(2sin(x/2)cos(x/2))/(1+1-2sin^2(x/2))
= -i(sin(x/2)cos(x/2))/(cos^2(x/2))
= -isin(x/2)/cos(x/2)
= -itan(x/2)

In summary: times top and bottom by conjugate of the denominator, i.e. realising it.

3. ## Re: Help with Complex Number Question!

Originally Posted by yleming
Not sure how to do this question:

If z = cos θ + isin θ, show that (1-z)/(1+z) = -itan(θ/2)

Any help would be appreciated. Thanks!
\begin{align*} \text{LHS } &= \frac{1-z}{1+z} \\ &= \frac{z^{1/2}(z^{-1/2}-z^{1/2})}{z^{1/2}(z^{-1/2}+z^{1/2})} \\ &= \frac{z^{-1/2}-z^{1/2}}{z^{-1/2}+z^{1/2}} \\ \end{align*}

$\text{Now observe that since } |z| = 1, \text{ we have } z^{-1} = \overline{z} \Longrightarrow (z^{-1})^{1/2} = (\overline{z})^{1/2} \Longrightarrow z^{-1/2} = \overline{z^{1/2}}.$

\begin{align*} \therefore \text{LHS } &= \frac{\overline{z^{1/2}}-z^{1/2}}{\overline{z^{1/2}}+z^{1/2}} \\ &= \frac{-2i\text{Im}(z^{1/2})}{2\text{Re}(z^{1/2})} \\ &= -i\frac{\sin{\theta/2}}{\cos{\theta/2}} \\ &= -i\tan\theta/2 \\ &= \text{RHS} \end{align*}

$\text{QED}$

Thanks!

5. ## Re: Help with Complex Number Question!

Originally Posted by aa180
\begin{align*} \text{LHS } &= \frac{1-z}{1+z} \\ &= \frac{z^{1/2}(z^{-1/2}-z^{1/2})}{z^{1/2}(z^{-1/2}+z^{1/2})} \\ &= \frac{z^{-1/2}-z^{1/2}}{z^{-1/2}+z^{1/2}} \\ \end{align*}

$\text{Now observe that since } |z| = 1, \text{ we have } z^{-1} = \overline{z} \Longrightarrow (z^{-1})^{1/2} = (\overline{z})^{1/2} \Longrightarrow z^{-1/2} = \overline{z^{1/2}}.$

\begin{align*} \therefore \text{LHS } &= \frac{\overline{z^{1/2}}-z^{1/2}}{\overline{z^{1/2}}+z^{1/2}} \\ &= \frac{-2i\text{Im}(z^{1/2})}{2\text{Re}(z^{1/2})} \\ &= -i\frac{\sin{\theta/2}}{\cos{\theta/2}} \\ &= -i\tan\theta/2 \\ &= \text{RHS} \end{align*}

$\text{QED}$

How did you think of that? Like i would have done like pikachu but what made you go do the qtn in that way?

6. ## Re: Help with Complex Number Question!

Originally Posted by HeroWise
How did you think of that? Like i would have done like pikachu but what made you go do the qtn in that way?
Just a trick I learnt at uni

8. ## Re: Help with Complex Number Question!

Originally Posted by HeroWise
Lol, just maximize your exposure to various resources as well as students/teachers who are adept in this subject and you'll be fine

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