Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises - Page 64: 83

$z=a+bi$ So, $z\times\frac{}{z}=a^{2}+b^{2}$ is a real number

$z=a+bi$ Find the conjugate of the complex numbers by changing the sign of their imaginary part: $\frac{}{z}=a-bi$ $z\times\frac{}{z}=(a+bi)+(a-bi)=a^{2}-(bi)^{2}$ We have: $i^{2}=-1$, So $z\times\frac{}{z}= a^{2}-b^{2}\times(-1)=a^{2}+b^{2}$ Therefore, $z\times\frac{}{z}$ is a real number