# Math Help - complex number and trigo Urgently

1. ## complex number and trigo Urgently

given that z =cosθ +jsinθ
((z²-1)/(z²+1))=jtanθ
Can someone prove it to mi pls. THX

2. can any1 pls help mi????
i nid help

3. Originally Posted by juzo
given that z =cosθ +jsinθ
((z²-1)/(z²+1))=jtanθ
Can someone prove it to mi pls. THX
Substitute for z in numerator and denominator and expand.

Substitute $\cos^2 \theta - 1 = - \sin^2 \theta$ and $1 - \sin^2 \theta = \cos^2 \theta$.

Simplify:

$\frac{- \sin^2 \theta + j \cos \theta \sin \theta}{\cos^2 \theta + j \cos \theta \sin \theta} = \frac{\sin \theta (- \sin \theta + j \cos \theta)}{\cos \theta (\cos \theta + j \sin \theta)}$

$= \frac{\sin \theta (- \sin \theta + j \cos \theta)}{\cos \theta (\cos \theta + j \sin \theta)}$

$= \frac{j \sin \theta \left(- \frac{\sin \theta}{j} + \cos \theta\right)}{\cos \theta (\cos \theta + j \sin \theta)}$

$= \frac{j \sin \theta \left(j \sin \theta + \cos \theta\right)}{\cos \theta (\cos \theta + j \sin \theta)} = \frac{j \sin \theta}{\cos \theta}$

4. Originally Posted by mr fantastic
Substitute for z in numerator and denominator and expand.

Substitute $\cos^2 \theta - 1 = - \sin^2 \theta$ and $1 - \sin^2 \theta = \cos^2 \theta$.

Simplify:

$\frac{- \sin^2 \theta + j \cos \theta \sin \theta}{\cos^2 \theta + j \cos \theta \sin \theta} = \frac{\sin \theta (- \sin \theta + j \cos \theta)}{\cos \theta (\cos \theta + j \sin \theta)}$

$= \frac{\sin \theta (- \sin \theta + j \cos \theta)}{\cos \theta (\cos \theta + j \sin \theta)}$

$= \frac{j \sin \theta \left(- \frac{\sin \theta}{j} + \cos \theta\right)}{\cos \theta (\cos \theta + j \sin \theta)}$

$= \frac{j \sin \theta \left(j \sin \theta + \cos \theta\right)}{\cos \theta (\cos \theta + j \sin \theta)} = \frac{j \sin \theta}{\cos \theta}$
Sry, i dun get the first line. can u show a bit more of the steps of the first part

5. Originally Posted by juzo
Sry, i dun get the first line. can u show a bit more of the steps of the first part
Did you substitute $z = \cos \theta + j \sin \theta$ ?

Did you expand?

Did you substitute and ?

6. Given that z=cosθ+jsinθ show that
((cos2θ+sin2θ-1)/(cos2θ+sin2θ+1))
den i dunno sub wht inside to get correct

7. Hello, juzo!

I have a looong proof . . . hope someone finds a shorter one!

Given: . $z\: =\:\cos\theta +j\sin\theta$

Prove: . $\frac{z^2-1}{z^2+1} \:=\:j\tan\theta$

Using DeMoivre's Theorem, we have:

. . $z^2 - 1 \;=\;(\cos\theta + j\sin\theta)^2 - 1 \;=\;\cos2\theta + j\sin2\theta - 1$

. . $z^2 + 1 \;=\;(\cos\theta + j\sin\theta)^2 + 1 \;=\;\cos2\theta + j\sin2\theta - 1$

So we have: . $\frac{z^2-1}{z^2+1} \;=\;\frac{(\cos2\theta-1) + j\sin2\theta}{(\cos2\theta + 1) + j\sin2\theta}$

Rationalize: . $\frac{(\cos2\theta-1) + j\sin2\theta}{(\cos2\theta + 1) + j\sin2\theta} \cdot {\color{blue}\frac{(\cos2\theta+1) - j\sin2\theta}{(\cos2\theta + 1) - j\sin2\theta}}$

. . $= \;\frac{\overbrace{(\cos^2\!2\theta - 1)}^{-\sin^2\!2\theta} - j\sin2\theta(\cos2\theta-1) + j\sin2\theta(\cos2\theta+1) \:\overbrace{- j^2\sin^2\!2\theta}^{+\sin^22\theta}}{(\cos2\theta +1)^2 - j^2\sin^2\!2\theta}$

. . $= \;\frac{-j\sin2\theta\cos2\theta + j\sin2\theta + j\sin2\theta\cos2\theta + j\sin2\theta}{\cos^2\!2\theta + 2\cos2\theta + 1 + \sin^2\!2\theta}$

. . $= \;\frac{2j\sin2\theta}{2 + 2\cos2\theta} \;=\;\frac{j\sin2\theta}{1 + \cos2\theta} \;=\;\frac{j\cdot2\sin\theta\cos\theta}{2\cos^2\!\ theta}$

. . $= \;\frac{j\sin\theta}{\cos\theta} \;=\;j\tan\theta$

8. Originally Posted by juzo
Given that z=cosθ+jsinθ show that
((cos2θ+sin2θ-1)/(cos2θ+sin2θ+1))
den i dunno sub wht inside to get correct
I was expecting to see $z^2 = (\cos \theta + j \sin \theta)^2 = \cos^2 \theta + 2 j \cos \theta \sin \theta - \sin^2 \theta$

By the way, please try to use proper English, NOT den, dunno, wht, mi pls etc. (which I personally find very annoying).

9. Let $z=\cos \theta + i \sin \theta$ this means $z = e^{i\theta}$.

Thus, $\frac{z^2-1}{z^2+1} = \frac{e^{2i\theta}-1}{e^{2i\theta}+1} = \frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}} = \frac{\sinh (i\theta)}{\cosh (i\theta)} = \frac{i\sin \theta}{\cos \theta} = i\tan \theta$

10. Originally Posted by ThePerfectHacker
Let $z=\cos \theta + i \sin \theta$ this means $z = e^{i\theta}$.

Thus, $\frac{z^2-1}{z^2+1} = \frac{e^{2i\theta}-1}{e^{2i\theta}+1} = \frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}} = \frac{\sinh (i\theta)}{\cosh (i\theta)} = \frac{i\sin \theta}{\cos \theta} = i\tan \theta$
Good post, but it's likely that juzo won't have done complex analysis to this level.

Like Mr Fantastic says, substitute $z = \cos(\theta) + j\sin(\theta)$ into the equation.

So $\frac{z^2-1}{z^2+1} = \frac{(\cos(\theta) + j\sin(\theta))^2 - 1}{(\cos(\theta) + j\sin(\theta))^2 + 1} = \frac{\cos^2(\theta) + 2j\cos(\theta)\sin(\theta) - \sin^2(\theta) - 1}{\cos^2(\theta) + 2j\cos(\theta)\sin(\theta) - \sin^2(\theta) + 1}.$

Then the rest of Mr Fantastic's work follows.

Hope that helped.