(iii) ∣z∣2=z zˉ|z|^2 = z\,\bar{z}∣z∣2=zzˉ Solution Let z=a+ib, a,b∈R.zˉ=a−ibzzˉ=(a+ib)(a−ib)=a2+b2But ∣z∣=a2+b2.⇒∣z∣2=a2+b2=zzˉ\begin{aligned} & \boxed{\text{Let } z=a+ib,\ a,b\in\mathbb{R}.} \\ \\ & \bar{z}=a-ib \\ \\ & z\bar{z}=(a+ib)(a-ib)=a^2+b^2 \\ \\ & \boxed{\text{But } |z|=\sqrt{a^2+b^2}.} \\ \\ & \Rightarrow |z|^2=a^2+b^2=z\bar{z} \end{aligned}Let z=a+ib, a,b∈R.zˉ=a−ibzzˉ=(a+ib)(a−ib)=a2+b2But ∣z∣=a2+b2.⇒∣z∣2=a2+b2=zzˉ