# Complex Variables Algebra Solutions / Argument/Modulus

1. Feb 26, 2012

### gbu

1. The problem statement, all variables and given/known data

Solve for a, $a \in \mathbb{C}$

$\frac{2\ln(a^2 - 1)}{\pi i} = 1$

2. Relevant equations
N/A.

3. The attempt at a solution

Reorganizing the equation.
$2\log(a^2 - 1) = \pi i$

Last edited: Feb 26, 2012
2. Feb 26, 2012

### fauboca

so a = x+yi. Then (x+yi)^2-1 = (x^2-y^2-1)+2xyi. What is the modulus of that?

Put your i's together as well.

3. Feb 26, 2012

### gbu

It's

$\sqrt{(4x^2 y^2 + (x^2-y^2-1)^2)}$

Which I suppose gets me to an answer, but its certainly not a pretty one. Wolfram Alpha gives a very simple answer to the question (a^2 = sqrt(1+i))

4. Feb 26, 2012

### fauboca

There may be a better way to do this one. Let me think.

5. Feb 26, 2012

### fauboca

Separate to log = pi i/2 and raise use the inverse function e.

6. Feb 26, 2012

### Dick

Look at what the parts of
$2\ln(|a^2 - 1|) + 2 i \arg(a^2 - 1) = \pi i$
are telling you. Equate real and imaginary parts, so $\ln(|a^2-1|)=0$ and $\arg(a^2 - 1)= \frac{\pi}{2}$. Without solving any equations and just thinking geometrically, what do those two things tell you about $a^2-1$?