# Another complex number

• Dec 10th 2008, 04:50 AM
Another complex number
If $\displaystyle z=x+yi$ and $\displaystyle z^2=a+bi$ , x,y,a,b are all real numbers . Prove that $\displaystyle 2x^2=\sqrt{(a^2+b^2)}+a$
By solving the equation $\displaystyle z^4+6z^2+25=0$ for $\displaystyle z^2$ or otherwise , express each root of the equation in the form of $\displaystyle x+yi$
• Dec 10th 2008, 11:20 AM
running-gag
Quote:

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

Hi

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

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

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

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

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

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

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

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