The question:

$\displaystyle |\sqrt{w^2 - 4} + w| = 2$ such that w is a complex number

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

Any assistance would be great.

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- Mar 18th 2011, 08:30 PMGlitchSimplify complex equation
**The question:**

$\displaystyle |\sqrt{w^2 - 4} + w| = 2$ such that w is a complex number

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

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

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

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

Now try to solve for $\displaystyle \displaystyle w$... - Mar 19th 2011, 02:48 AMmr fantastic
- Mar 19th 2011, 04:25 AMGlitch
- Mar 19th 2011, 12:45 PMawkward
- Mar 19th 2011, 04:21 PMGlitch
- Mar 19th 2011, 04:33 PMProve It
Because on the unit circle, $\displaystyle \displaystyle |z|^2 = 1 \implies z\overline{z} = 1 \implies \overline{z} = \frac{1}{z}$.

- Mar 19th 2011, 07:02 PMGlitch
Oh, right. >_<

I need stronger coffee. - Mar 19th 2011, 10:30 PMGlitch
OK, so I've worked out what you've done so far. This is what I've done:

Let z = x + iy

$\displaystyle (x + iy + x - iy)^2 - 2$

$\displaystyle (2x)^2 - 2$

$\displaystyle 4x^2 - 2$

So unless I'm mistaken, this is a parabola. I'm fairly sure this isn't the correct solution. :/ - Mar 20th 2011, 04:17 AMawkward
- Mar 20th 2011, 08:09 PMGlitch
Ahh, thanks.