1. ## 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.

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

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

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

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

3. 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?)

4. 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]

5. 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...

6. 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 $z^{-1} = \overline{z}$.

7. Because on the unit circle, $\displaystyle |z|^2 = 1 \implies z\overline{z} = 1 \implies \overline{z} = \frac{1}{z}$.

8. Oh, right. >_<

I need stronger coffee.

9. 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. :/

10. 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 $4x^2 - 2$?

11. Ahh, thanks.