First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.1

Solution

Prove each identity.

Hint

Write z = a + bi and z̄ = a - bi.

Then compute:

(z + z̄)/2
(z - z̄)/(2i)
z z̄

(i) $\frac{z+\bar{z}}{2}=\operatorname{Re}(z)$

\begin{aligned} & \boxed{\text{Let } z=a+ib, \ a,b\in\mathbb{R}.} \\ \\ & \boxed{\text{Then } \bar{z}=a-ib.} \\ \\ & z+\bar{z}=(a+ib)+(a-ib)=2a \\ \\ & \frac{z+\bar{z}}{2} = \frac{2a}{2}=a=\operatorname{Re}(z) \end{aligned}

(ii) $\frac{z-\bar{z}}{2i}=\operatorname{Im}(z)$

\begin{aligned} & \boxed{\text{Let } z=a+ib, \ a,b\in\mathbb{R}.} \\ \\ & \boxed{\text{Then } \bar{z}=a-ib.} \\ \\ & z-\bar{z}=(a+ib)-(a-ib)=2ib \\ \\ & \frac{z-\bar{z}}{2i} = \frac{2ib}{2i}=b=\operatorname{Im}(z) \end{aligned}

(iii) $|z|^2 = z\,\bar{z}$

\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}

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