Exercise 1.2 — Question 5

Key Concepts Covered

This question covers basic operations on complex numbers (SLO M-11-A-04), the complex conjugate (SLO M-11-A-05), and the modulus of a complex number (SLO M-11-A-06).

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Background Definitions

Complex Conjugate: For $z=a+bi$, the conjugate is $\bar{z}=a-bi$.

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

Purely Real: $z$ is purely real if $b=0$, i.e. $z=\bar{z}$.

Purely Imaginary: $z$ is purely imaginary if $a=0$ and $b\ne 0$, i.e. $z+\bar{z}=0$ (equivalently $z=-\bar{z}$).

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

Interpretation: Multiplying a complex number by $i$ swaps its real and imaginary parts, negating the new real part. Geometrically, this corresponds to a $90°$ counter-clockwise rotation.

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Part (ii): Condition for $z$ to be Purely Real (using conjugate)

For $z=a+bi$:

• $\bar{z}=a-bi$
• $z=\bar{z}⟺a+bi=a-bi⟺2bi=0⟺b=0$

$\boxed{z\text{ is purely real}⟺z=\bar{z}}$

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Part (iii): Modulus and $z\cdot \bar{z}$

For $z=a+bi$: $z\cdot \bar{z}=(a+bi)(a-bi)=a^2-(bi{)}^2=a^2+b^2$

Since $|z|=\sqrt{a^2+b^2}$, we have: $z\cdot \bar{z}=|z{|}^2$

This is a key identity used to divide complex numbers.

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

For $z=a+bi$:

• $z$ is purely imaginary $⟺a=0$ and $b\ne 0$
• In terms of conjugate: $z+\bar{z}=(a+bi)+(a-bi)=2a=0⟺a=0$

$\boxed{z\text{ is purely imaginary}⟺z+\bar{z}=0,z\ne 0}$