Exercise 1.2 — Question 3

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

Key Definitions

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

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

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Standard Results (Q3 Type)

1. Multiplication by $i$

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

Therefore:

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

Conclusion: Multiplying by $i$ rotates the complex number — the real and imaginary parts swap with a sign change on the real part.

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2. Sum and Difference with Conjugate

For $z=a+bi$ and $\bar{z}=a-bi$:

$z+\bar{z}=(a+bi)+(a-bi)=2a=2\mathrm{Re}(z)$

$z-\bar{z}=(a+bi)-(a-bi)=2bi=2i\mathrm{Im}(z)$

Key results: $\mathrm{Re}(z)=\frac{z+\bar{z}}{2},\mathrm{Im}(z)=\frac{z-\bar{z}}{2i}$

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3. Product with Conjugate (Modulus)

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

This is always a non-negative real number.

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4. Condition for $z$ to be Purely Real

$z$ is purely real $⟺$ $\mathrm{Im}(z)=0$ $⟺$ $b=0$.

In terms of conjugate: $z=\bar{z}$, since $a+bi=a-bi\Rightarrow 2bi=0\Rightarrow b=0$.

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5. Condition for $z$ to be Purely Imaginary

$z$ is purely imaginary $⟺$ $\mathrm{Re}(z)=0$ $⟺$ $a=0$.

In terms of conjugate: $z=-\bar{z}$, since $a+bi=-(a-bi)\Rightarrow 2a=0\Rightarrow a=0$.

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Summary Table