Show that the product of an equation is equal to the square of its modulus.

Question

How would I do this?

Show that the product of $\bar {z} = r[\cos⁡(-θ)+i \sin(-θ)]$ and $z=r(\cosθ+i\sinθ)$ is equal to the square of the modulus.

Answer 1

$$(a+ib)(a-ib)=a^2+b^2=(\sqrt{a^2+b^2})^2=|z|^2.$$

Answer 2

Well \begin{align} z \bar{z} &= r(\cos(-\theta)+i \sin (-\theta))r(\cos(\theta)+i \sin (\theta)) \\ &= r^2(\cos(\theta)-i \sin (\theta))(\cos(\theta)+i \sin (\theta)) \\ &= r^2(\cos^2 \theta -i \sin \theta \cos \theta \ldots) \\ \end{align}

Can you finish?

Answer 3

Hint: if $z = r \cos \theta + i \sin \theta$ and $\bar z = r \cos (-\theta) + i \sin (-\theta)$, then $$z \bar z= r^2 (\cos (\theta - \theta) + i \sin (\theta - \theta))$$ but $\cos (\theta -\theta) = ?$ and $\sin (\theta -\theta) = ?$