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

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[1-itan(alfa)]

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

Thanks

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- Sep 27th 2007, 05:06 AM #1

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- Sep 27th 2007, 05:30 AM #2
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

- Sep 27th 2007, 05:41 AM #3
Recall that $\displaystyle \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$ so that $\displaystyle \frac{1}{{1 - i\tan (\alpha )}} = \frac{{1 + i\tan (\alpha )}}{{\sec ^2 (\alpha )}}$.

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

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

- Sep 27th 2007, 07:56 AM #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}$

- Sep 27th 2007, 08:24 AM #5