1.2 Q-2

Exercise 1.2 — Question 2

This exercise covers basic operations on complex numbers, including multiplication by $i$ and conditions for a complex number to be real or purely imaginary.

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Key Concepts

Multiplication by $i$:

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

So:

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

Multiplying a complex number by $i$ rotates it by $90°$ in the complex plane.

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Condition for $z$ to be purely real:

$z=a+bi$ is purely real $⟺b=0⟺z=\bar{z}$

Proof: If $z=\bar{z}$, then $a+bi=a-bi\Rightarrow 2bi=0\Rightarrow b=0$.

Condition for $z$ to be purely imaginary:

$z=a+bi$ is purely imaginary $⟺a=0⟺z=-\bar{z}$

Proof: If $z=-\bar{z}$, then $a+bi=-(a-bi)=-a+bi\Rightarrow 2a=0\Rightarrow a=0$.

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Worked Examples

Example 1: If $z=3+4i$, find $iz$ and identify its real and imaginary parts.

$iz=i(3+4i)=3i+4i^2=3i-4=-4+3i$

$\mathrm{Re}(iz)=-4=-\mathrm{Im}(z)$ ✓
$\mathrm{Im}(iz)=3=\mathrm{Re}(z)$ ✓

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Example 2: Show that $z+\bar{z}$ is always real for any complex number $z=a+bi$.

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

Since $2a$ has no imaginary part, $z+\bar{z}$ is always real.

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Example 3: Show that $z-\bar{z}$ is always purely imaginary (or zero).

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

Since $2bi$ has no real part, $z-\bar{z}$ is always purely imaginary (or zero when $b=0$).

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