Complex Variables Algebra Solutions / Argument/Modulus

[gbu]

Homework Statement

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

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

Homework Equations

N/A.

The Attempt at a Solution

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

 

[fauboca]

Homework Statement

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

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

Homework Equations

N/A.

The Attempt at a Solution

Reorganizing the equation.
$2\log(a^2 - 1) = \pi i$
Expanding the logarithm.
$2\ln(|a^2 - 1|) + i \textrm{arg}(a^2 - 1) = \pi i$I think what I'm stuck on is that I don't know how to evaluate my length/argument of an arbitrary complex variable like that. I know how to solve them if I'm given a value of a (a = x + iy, then $|a| = \sqrt(x^2 + y^2)$ and $arg(a) = \tan^{-1} \frac{y}{x}$), but without the value of a I'm not sure where to go.

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.

 

[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))

 

[fauboca]

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))

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

 

[fauboca]

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

 

[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 $a^2-1$?