Another complex number

Quote:

Originally Posted by mathaddict

If $z=x+yi$ and $z^2=a+bi$ , x,y,a,b are all real numbers . Prove that $2x^2=\sqrt{(a^2+b^2)}+a$
By solving the equation $z^4+6z^2+25=0$ for $z^2$ or otherwise , express each root of the equation in the form of $x+yi$

Hi

If $z=x+yi$ then
$z^2=x^2-y^2+2xyi$

Then
(i) $x^2-y^2=a$
(ii) $2xy=b$ then $y=\frac{b}{2x}$

Substituting y in (i)
$x^2 - \frac{b^2}{(2x)^2}=a$

$(2x^2)^2 - 2a (2x^2) - b^2=0$

Reduced discriminant is $a^2 + b^2$

The only positive solution is therefore $2x^2=\sqrt{a^2+b^2}+a$

Equation $z^4+6z^2+25=0$ for $z^2$
gives $z^2 = -3 + 4i$ or $z^2 = -3 - 4i$

It is just an application with a=-3, b=4 and then a=-3 and b=-4