# Deriving inverse tans from complex product?

1. Apr 15, 2007

### rsnd

I am supposed to derive $4\,\arctan \left( 1/5 \right) -\arctan \left( {\frac {1}{239}} \right)$ from the complex product of $\left( 1+i \right) \left( 5-i \right) ^{4}$ I do see how the argument of product of the complex expression is equal to pi/4 - 4 arctan(1/5) but I am totally lost. SO how am I supposed to approach this? Thanks heaps in advance.

2. Apr 15, 2007

### Gib Z

Well You have to equate to long way and the short way :D

You already established that the argument to the complex product is $\frac{\pi}{4} -4\arctan \frac{1}{5}$

What you had to do next is have the pleasure of EXPANDING YOUR PRODUCT. Binomial theorem is just too much of a hassle this time. Write the product as
$$(5-i)^{2^2}\cdot(1+i)$$. So just square 5-i, 2 times, and multiply that by 1+i.

FUN FUN, I did it for you :P You get : 956-4i . The argument of this is arctan (-4/956)= - arctan (1/239) . Isn't that interesting?

3. Apr 16, 2007

### rsnd

And here I thought I'd never have to multiply the hard way after learning the polar form. Thanks heaps again =)

4. Apr 16, 2007