Given two numbers $u$ and $v$, prove there exist (possibly complex) numbers $y$ and $z$ such that $y+z=u$ and $yz=v$.

Question

Given two numbers $u$ and $v$, prove there exist (possibly complex) numbers $y$ and $z$ such that $y+z=u$ and $yz=v$.

This is an statement Rotman mentions in many of his books in order to prove the cubic formula by Tartaglia, but I cannot figure out how to prove this step. Thanks.

Answer 1

You have to solve $y+\color{red}z=u$ and $y\color{red}z=v$ for $y$ and $\color{red}z$, given $u$ and $v$.

To do so, isolate one of the variables:

substitute $y=u-\color{red}z$ (from the first equation) into the second equation;

this yields $y\color{red}{(u-y)}=v$.

The equation $y^2-uy+v=0$ can be solved in $\mathbb C$ using the quadratic formula:

$$y=\dfrac{u\pm\sqrt{u^2-4v}}2.$$

You can then find $\color{red}z$ using $\color{red}z=u-y$ and check that $y+\color{red}z=u$ and $y\color{red}z=v.$

Answer 2

Consider the quadratic equation: $$z^2-uz+v=0.$$ This equation has two $x$ and $y$ (potentially equal). It is easy to verify that $ u=x+y$ and $v=xy$.