Multiplying a complex number by its conjugate

In the same way that multiplying a bracket involving a surd by the conjugate of the bracket will produce a rational number, the same is try for multiplying a complex number by the complex conjugate.

Multiplying a complex number by its conjugate will always result in a real number.

Proof

• Let $z=x+yi$. $x$ and $y$ are real.
• $z^*=x-yi$
• $(x+yi)(x-yi)=x^2+xyi-xyi-y^2i^2$
• $(x+yi)(x-yi)=x^2-y^2i^2$
• $(x+yi)(x-yi)=x^2-y^2(-1)$
• $(x+yi)(x-yi)=x^2+y^2$
• $x$ and $y$ are real, so $x^2+y^2$ is real.

Example: find $z$ such that $3z+2z^*=5+2i$

• Let $z=x+yi$
• $z^*=x-yi$
• $3(x+yi)+2(x-yi)=5+2i$
• $3x+3yi+2x-2yi=5+2i$
• $5x+yi=5+2i$
• $5x=5$, $x=1$
• $1y=2$, $y=2$
• $z=x+yi$
• $z=1+2i$