Exercise 1.2 — Question 9

Key Concepts Reviewed

This question applies three core ideas about complex numbers:

1. Basic operations on complex numbers (SLO M-11-A-04)
2. Complex conjugate: $\bar{z}=a-ib$ for $z=a+ib$ (SLO M-11-A-05)
3. Modulus: $|z|=\sqrt{a^2+b^2}$ for $z=a+ib$ (SLO M-11-A-06)

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Part (i): Find $\mathrm{Re}(iz)$ and $\mathrm{Im}(iz)$

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): Condition for $z$ to be Purely Real

A complex number $z=a+bi$ is purely real when $b=0$.

In terms of the conjugate $\bar{z}=a-bi$:

$z=\bar{z}⟺a+bi=a-bi⟺2bi=0⟺b=0$

Conclusion: $z$ is purely real $⟺z=\bar{z}$.

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Part (iii): Condition for $z$ to be Purely Imaginary

A complex number $z=a+bi$ is purely imaginary when $a=0$ (and $b\ne 0$).

In terms of the conjugate:

$z=-\bar{z}⟺a+bi=-(a-bi)=-a+bi⟺2a=0⟺a=0$

Conclusion: $z$ is purely imaginary $⟺z=-\bar{z}$.

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Part (iv): Modulus and the Product $z\cdot \bar{z}$

For $z=a+bi$, the modulus is: $|z|=\sqrt{a^2+b^2}$

The product of $z$ and its conjugate: $z\cdot \bar{z}=(a+bi)(a-bi)=a^2-(bi{)}^2=a^2+b^2=|z{|}^2$

This gives the useful identity: $z\cdot \bar{z}=|z{|}^2$

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Summary Table