complex prove

por **stuart clark** » Dom Abr 15, 2012 13:05

Prove that $\tan \left(i.\ln\left(\frac{a-ib}{a+ib}\right)\right) = -\frac{2ab}{a^2-b^2}$

por **fraol** » Sáb Abr 21, 2012 16:28

stuart clark escreveu:Prove that $\tan \left(i.\ln\left(\frac{a-ib}{a+ib}\right)\right) = -\frac{2ab}{a^2-b^2}$

Let

be a complex number such that:

z = a + bi

and $z = r(cos \theta + i sin \theta)$ .

So: $a + bi = r(cos \theta + i sin \theta)$ . Then we have:

$a = rcos \theta$ and $b = r sin \theta$ .

Now, with results above, let's do some algebraic manipulation with the expression inside natural log and use the Euler identity $e^{i\theta} = cos \theta + i sin \theta$ :

$\frac{a - bi}{a + bi} = \frac{rcos \theta - i r sin \theta}{rcos \theta + i r sin \theta} = \frac{cos \theta - i sin \theta}{cos \theta + i sin \theta} = \frac{e^{-i\theta}}{e^{i\theta}} = e^{-2i\theta}$ .

Returning to the original expression and applying the last result we get:

$\tan \left(i.\ln\left(\frac{a-ib}{a+ib}\right)\right) = tan \left(i.\ln e^{-2i\theta} \right) = tan \left( -2i^2\theta \right) = tan \left( 2\theta \right)$ .

From trigonometry, we have:

$tan \left( 2\theta \right) = \frac{2 tan \theta}{1 - tan^2 \theta} = \frac{2cos \theta sin \theta}{cos^2 \theta - sin^2 \theta }$ .

Replacing cos

and

in this last expression:

$\frac{2cos \theta sin \theta}{cos^2 \theta - sin^2 \theta } = \frac{2 \frac{a}{r} . \frac{b}{r} } {\frac{a^2}{r^2} - \frac{b^2}{r^2}} = \frac{2ab}{a^2 - b^2}$ ,

that is the desired result (note that this solution doesn't contains the negative sign ).

.

por **stuart clark** » Qua Mai 02, 2012 01:07

Thanks fraol

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