Theory of Intelligence
11TH•Mathematics•Complex Numbers•Question 3

Solution

Prove that $z=\bar{z}$ if and only if $z$ is real.

Solution

Let z=a+ib, where a,b∈R.Then the conjugate is zˉ=a−ib.(\Rightarrow) If z=zˉ:a+ib=a−ib⇒ib=−ib⇒2ib=0⇒b=0So z=a+0i=a, which is real.(\Leftarrow) If z is real, then z=a+0i.⇒zˉ=a−0i=a=z∴ z=zˉ  ⟺  z is real.\begin{aligned} & \boxed{\text{Let } z=a+ib, \text{ where } a,b\in\mathbb{R}.} \\ \\ & \boxed{\text{Then the conjugate is } \bar{z}=a-ib.} \\ \\ & \boxed{\text{(\Rightarrow) If } z=\bar{z}:} \\ \\ & a+ib = a-ib \\ & \Rightarrow ib = -ib \\ & \Rightarrow 2ib = 0 \\ & \Rightarrow b = 0 \\ \\ & \boxed{\text{So } z=a+0i=a, \text{ which is real.}} \\ \\ & \boxed{\text{(\Leftarrow) If } z \text{ is real, then } z=a+0i.} \\ \\ & \Rightarrow \bar{z}=a-0i=a=z \\ \\ & \boxed{\therefore\ z=\bar{z} \iff z \text{ is real.}} \end{aligned}

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