1. ## Complex numbers

I don't know where to post this, so I'll put it here.
If $z=x+yi$ and $z^2=a+ib$ and a, b, x, and y are real, prove that
$2x^2=\sqrt{a^2+b^2}+a$

How do i go about doing this? Other than that
$a=x^2-y^2$ and $b=2xy$
Thanks

2. Originally Posted by arze
I don't know where to post this, so I'll put it here.
If $z=x+yi$ and $z^2=a+ib$ and a, b, x, and y are real, prove that
$2x^2=\sqrt{a^2+b^2}+a$

How do i go about doing this? Other than that
$a=x^2-y^2$ and $b=2xy$
Thanks
you have:

$y=\frac{b}{2x}$

substitute this into the other equation to get:

$x^2-\frac{b^2}{4x^2}-a=0$

Multiply this through by $x^2$ to get a quadratic in $x^2$, and solve for $x^2$ using the quadratic formula and discard the negative solution.

CB

3. so we have $4x^2-ax^2-b^2=0$
$x^2=\frac{a\pm\sqrt{a^2+16b^2}}{8}$
then $2x^2=\frac{a+\sqrt{a^2+16b^2}}{4}$
correct?
how does this become $2x^2=\sqrt{a^2+b^2}+a$?

4. Originally Posted by arze
so we have $4x^2-ax^2-b^2=0$
$x^2=\frac{a\pm\sqrt{a^2+16b^2}}{8}$
then $2x^2=\frac{a+\sqrt{a^2+16b^2}}{4}$
correct?
how does this become $2x^2=\sqrt{a^2+b^2}+a$?
You get:

$(x^2)^2-a(x^2)-\frac{b^2}{4}=0$

so:

$x^2=\frac{a\pm\sqrt{a^2+\frac{4b^2}{4}}}{2}=\frac{ a\pm\sqrt{a^2+b^2}}{2}$

Now as $x^2$ is intrinsically positive we discard the negative root in the above to get:

$x^2=\frac{a+\sqrt{a^2+b^2}}{2}$

CB