1.2 Q-8

Exercise 1.2 — Question 8

This question covers basic operations on complex numbers, the complex conjugate, and properties of the modulus.

Problem

If $z=a+bi$, show that:

(i) $\mathrm{Re}(iz)=-\mathrm{Im}(z)$ and $\mathrm{Im}(iz)=\mathrm{Re}(z)$

(ii) $z$ is purely real if and only if $z=\bar{z}$

(iii) $z$ is purely imaginary if and only if $z=-\bar{z}$

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Solution

Part (i): Real and Imaginary Parts of $iz$

Let $z=a+bi$. Multiply by $i$:

$iz=i(a+bi)=ai+b{i}^2=ai+b(-1)=-b+ai$

Therefore: $\mathrm{Re}(iz)=-b=-\mathrm{Im}(z)$ $\mathrm{Im}(iz)=a=\mathrm{Re}(z)$

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Part (ii): $z$ is Purely Real $⟺$ $z=\bar{z}$

Recall $\bar{z}=a-bi$.

($\Rightarrow$) If $z$ is purely real, then $b=0$, so $z=a$ and $\bar{z}=a$. Hence $z=\bar{z}$.

($\Leftarrow$) If $z=\bar{z}$, then: $a+bi=a-bi⟹2bi=0⟹b=0$ So $z$ is purely real. $■$

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Part (iii): $z$ is Purely Imaginary $⟺$ $z=-\bar{z}$

Recall $\bar{z}=a-bi$, so $-\bar{z}=-a+bi$.

($\Rightarrow$) If $z$ is purely imaginary, then $a=0$, so $z=bi$ and $-\bar{z}=bi$. Hence $z=-\bar{z}$.

($\Leftarrow$) If $z=-\bar{z}$, then: $a+bi=-a+bi⟹2a=0⟹a=0$ So $z$ is purely imaginary. $■$