1.2 Q-7

Exercise 1.2 — Question 7

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

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Key Concepts

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

Modulus: $|z|=\sqrt{a^2+b^2}$, and $z\cdot \bar{z}=a^2+b^2=|z{|}^2$.

Multiplication by $i$: If $z=a+bi$, then $iz=i(a+bi)=ai+b{i}^2=-b+ai$ so $\mathrm{Re}(iz)=-\mathrm{Im}(z)$ and $\mathrm{Im}(iz)=\mathrm{Re}(z)$.

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

Let $z=a+bi$.

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

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

Interpretation: Multiplying a complex number by $i$ rotates it by $90°$ in the complex plane, swapping and negating the real and imaginary parts.

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Part (ii): Condition for $z$ to be purely real using $\bar{z}$

$z=a+bi$ is purely real $⟺$ $b=0$.

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

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

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Part (iii): Modulus and conjugate product

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

$\boxed{|z|=\sqrt{a^2+b^2},z\cdot \bar{z}=|z{|}^2}$

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