Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.1

Solution

Let z=a+ib, where a,bR.Then the conjugate is zˉ=aib.(\Rightarrow) If z=zˉ:a+ib=aibib=ib2ib=0b=0So z=a+0i=a, which is real.(\Leftarrow) If z is real, then z=a+0i.zˉ=a0i=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}

Hint

Let z = a + bi and compute z\u0304. Compare z = z\u0304.