1. ## complexe2

Hello i have 3 exercices i finished them and i need the last questions

Write in trigonometrical form the complex numbers:
[1+itan(alfa)]^4
------------
[1-itan(alfa)]

it's one number but i don't know how to use maths letters here

Thanks

2. Originally Posted by iceman1
Hello i have 3 exercices i finished them and i need the last questions

Write in trigonometrical form the complex numbers:
[1+itan(alfa)]^4
------------
[1-itan(alfa)]

it's one number but i don't know how to use maths letters here

Thanks
There is a "clever" way to start simplifying this, but let's just take things at face value here.

The first thing you want to do is to get rid of that "i" in the denominator:
$\displaystyle \frac{(1 + i~tan(\alpha))^4}{1 - i~tan(\alpha)} = \frac{(1 + i~tan(\alpha))^4}{1 - i~tan(\alpha)} \cdot \frac{1 + i~tan(\alpha)}{1 + i~tan(\alpha)}$

$\displaystyle = \frac{(1 + i~tan(\alpha))^5}{1 - i^2~tan^2(\alpha)} = \frac{(1 + i~tan(\alpha))^5}{1 + tan^2(\alpha)} = \frac{(1 + i~tan(\alpha))^5}{sec^2(\alpha)}$

Now expand the numerator using the binomial theorem (or if you are masochistic, expand it out term by term) and separate the real and imaginary parts.

-Dan

3. Recall that $\displaystyle \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$ so that $\displaystyle \frac{1}{{1 - i\tan (\alpha )}} = \frac{{1 + i\tan (\alpha )}}{{\sec ^2 (\alpha )}}$.

$\displaystyle \frac{{\left( {1 + i\tan (\alpha )} \right)^4 }}{{1 - i\tan (\alpha )}} = \frac{{\left( {1 + i\tan (\alpha )} \right)^5 }}{{\sec ^2 (\alpha )}}$

$\displaystyle \frac{{\left( {1 + i\tan (\alpha )} \right)^5 }}{{\sec ^2 (\alpha )}} = \frac{{\frac{{\left( {\cos (\alpha ) + i\sin (\alpha )} \right)^5 }}{{\cos ^5 (\alpha )}}}}{{\sec ^2 (\alpha )}} = \frac{{\cos (5\alpha ) + i\sin (5\alpha )}}{{\cos ^3 (\alpha )}}$

4. This is a postscript, resulting from a PM.
It seems that the problem is actually $\displaystyle \left[ {\frac{{1 + \tan (\alpha )}}{{1 - \tan (\alpha )}}} \right]^4$.
In which case;
$\displaystyle \begin{array}{rcl} \left[ {\frac{{1 + i\tan (\alpha )}}{{1 - i\tan (\alpha )}}} \right]^4 & = & \left[ {\frac{{\cos (\alpha ) + i\sin (\alpha )}}{{\cos (\alpha ) - i\sin (\alpha )}}} \right]^4 \\ & = & \left[ {\cos (2\alpha ) + i\sin (2\alpha )} \right]^4 \\ & = & \cos (8\alpha ) + i\sin (8\alpha ) \\ \end{array}$

5. Thanks