1.2 Q-4

Exercise 1.2 — Question 4

This exercise covers basic operations on complex numbers, properties of the complex conjugate, and the modulus of a complex number.

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Key Concepts

Multiplication by $i$:

If $z=a+bi$, then: $iz=i(a+bi)=ai+b{i}^2=ai-b=-b+ai$

So $\mathrm{Re}(iz)=-b=-\mathrm{Im}(z)$ and $\mathrm{Im}(iz)=a=\mathrm{Re}(z)$.

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Complex Conjugate:

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

• $z$ is purely real $⟺b=0⟺z=\bar{z}$
• $z$ is purely imaginary $⟺a=0⟺z=-\bar{z}$
• $z+\bar{z}=2a=2\mathrm{Re}(z)$
• $z-\bar{z}=2bi=2i\mathrm{Im}(z)$
• $z\cdot \bar{z}=a^2+b^2=|z{|}^2$

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Modulus (Absolute Value):

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

Properties:

• $|z|\ge 0$, and $|z|=0⟺z=0$
• $|z|=|\bar{z}|$
• $|z_1{z}_2|=|z_1||z_2|$
• $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$, $z_2\ne 0$

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Worked Examples

Example 1: If $z=3+4i$, find $iz$, $\bar{z}$, and $|z|$.

$iz=i(3+4i)=3i+4i^2=-4+3i$ $\bar{z}=3-4i$ $|z|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

Example 2: Show that $z\cdot \bar{z}=|z{|}^2$.

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

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