Algebra 2 - Adding and Subtracting Complex Numbers

Addition and Subtraction of Complex Numbers

When we add or subtract two complex numbers, we add or subtract the real number parts, then add or subtract their imaginary parts and express it in the standard form of a complex number.

Consider two complex numbers ${z}_{1}=a+ib$ and ${z}_{2}=c+id$.

• Addition: ${z}_{1}+{z}_{2}=\left(a+ib\right)+\left(c+id\right)=\left(a+c\right)+i\left(b+d\right)$
• Subtraction: ${z}_{1}-{z}_{2}=\left(a+ib\right)-\left(c+id\right)=\left(a-c\right)+i\left(b-d\right)$

Solved Examples

Example 1: Add $-5+3i$ and $15-8i$.

Solution: After grouping, we add the real and the imaginary parts separately.

$\left(-5+3i\right)+\left(15-8i\right)=$$\left(-5+15\right)+\left(3i-8i\right)=$$10-5i$

Example 2: Find the value of $\left(20-3i\right)+\left(15i\right)$.

Solution:

Example 3: Subtract $6-8i$ from $4+3i$.

Solution: $\left(4+3i\right)-\left(6-8i\right)=$$4+3i-6+8i=$$\left(4-6\right)+\left(3i+8i\right)=$$-2+11i$

Example 4: Simplify $\left(-11+35i\right)-\left(35i-11\right)$.

Solution: $\left(11+35i\right)-\left(35i-11\right)=$$11+35i-35i+11=$$\left(11+11\right)+\left(35i-35i\right)=$$22$

Example 5: Subtract $\left(2\sqrt{2}+7i\right)$ from $\left(7\sqrt{2}-5i\right)$.

Solution: $\left(7\sqrt{2}-5i\right)-\left(2\sqrt{2}+7i\right)=$$7\sqrt{2}-5i-2\sqrt{2}-7i=$$\left(7\sqrt{2}-2\sqrt{2}\right)-\left(5i+7i\right)=$$5\sqrt{2}-12i$.

Cheat Sheet

• To add or subtract two complex numbers, we add or subtract the corresponding real and imaginary parts.

Blunder Areas

• Subtracting ${z}_{1}$ from ${z}_{2}$ means computing ${z}_{2}-{z}_{1}$ and must not be confused with ${z}_{1}-{z}_{2}$.