# Simplify complex equation

• Mar 18th 2011, 08:30 PM
Glitch
Simplify complex equation
The question:
$|\sqrt{w^2 - 4} + w| = 2$ such that w is a complex number

I tried using the identity $|z|^2 = z\overline{z}$ to get a solution, but it appears to create a mess. :/

Any assistance would be great.
• Mar 18th 2011, 09:05 PM
Prove It
$\displaystyle |\sqrt{w^2 - 4} + w| = 2$

$\displaystyle |\sqrt{w^2 - 4} + w|^2 = 2^2$

$\displaystyle (\sqrt{w^2 - 4} + w)^2 = 4$.

Now try to solve for $\displaystyle w$...
• Mar 19th 2011, 02:48 AM
mr fantastic
Quote:

Originally Posted by Prove It
$\displaystyle |\sqrt{w^2 - 4} + w| = 2$

$\displaystyle |\sqrt{w^2 - 4} + w|^2 = 2^2$

$\displaystyle (\sqrt{w^2 - 4} + w)^2 = 4$.

Now try to solve for $\displaystyle w$...

I'm not sure this will work since $|a|^2 = a \overline{a}$, not a^2, when a is complex.

@OP: I'd be inclined to substitute w = x + iy and take it from there, messy as it's likely to be .... (Where has the question come from?)
• Mar 19th 2011, 04:25 AM
Glitch
Quote:

Originally Posted by mr fantastic
I'm not sure this will work since $|a|^2 = a \overline{a}$, not a^2, when a is complex.

@OP: I'd be inclined to substitute w = x + iy and take it from there, messy as it's likely to be .... (Where has the question come from?)

It's part of a complex analysis question:

Show that the function f defined by $f(z) = z^2 + z^{-2}$ maps the unit circle onto the interval [-2, 2]
• Mar 19th 2011, 12:45 PM
awkward
Quote:

Originally Posted by Glitch
It's part of a complex analysis question:

Show that the function f defined by $f(z) = z^2 + z^{-2}$ maps the unit circle onto the interval [-2, 2]

If $z$ is on the unit circle then $|z|^2 = z \overline{z} = 1$, so $z^{-1} = \overline{z}$.

Then $f(z) = z^2 + \overline{z}^2 = (z + \overline{z})^2 - 2 z \overline{z} = (z + \overline{z})^2 -2$

So...
• Mar 19th 2011, 04:21 PM
Glitch
Quote:

Originally Posted by awkward
If $z$ is on the unit circle then $|z|^2 = z \overline{z} = 1$, so $z^{-1} = \overline{z}$.

Sorry, I'm not sure how you came to the conclusion that $z^{-1} = \overline{z}$.
• Mar 19th 2011, 04:33 PM
Prove It
Because on the unit circle, $\displaystyle |z|^2 = 1 \implies z\overline{z} = 1 \implies \overline{z} = \frac{1}{z}$.
• Mar 19th 2011, 07:02 PM
Glitch
Oh, right. >_<

I need stronger coffee.
• Mar 19th 2011, 10:30 PM
Glitch
Quote:

Originally Posted by awkward
$(z + \overline{z})^2 -2$

OK, so I've worked out what you've done so far. This is what I've done:

Let z = x + iy

$(x + iy + x - iy)^2 - 2$
$(2x)^2 - 2$
$4x^2 - 2$

So unless I'm mistaken, this is a parabola. I'm fairly sure this isn't the correct solution. :/
• Mar 20th 2011, 04:17 AM
awkward
Quote:

Originally Posted by Glitch
OK, so I've worked out what you've done so far. This is what I've done:

Let z = x + iy

$(x + iy + x - iy)^2 - 2$
$(2x)^2 - 2$
$4x^2 - 2$

So unless I'm mistaken, this is a parabola. I'm fairly sure this isn't the correct solution. :/

Since (x,y) is on the unit circle, $-1 \leq x \leq 1$.

So then what is the range of $4x^2 - 2$?
• Mar 20th 2011, 08:09 PM
Glitch
Ahh, thanks.