Complex Number Arithmetic

This lesson covers the four basic arithmetic operations on complex numbers: addition, subtraction, multiplication, and division. Mastering these operations is essential, as they underpin every other topic in the complex numbers unit.

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Addition and Subtraction

To add or subtract complex numbers, combine the real parts and imaginary parts separately:

$(a + bi) + (c + di) = (a + c) + (b + d)i$(a+bi)+(c+di)=(a+c)+(b+d)i

$(a + bi) - (c + di) = (a - c) + (b - d)i$(a+bi)−(c+di)=(a−c)+(b−d)i

Worked Example 1: Compute $(3 + 2i) + (1 - 5i)$(3+2i)+(1−5i).

$(3 + 2i) + (1 - 5i) = (3 + 1) + (2 + (-5))i = 4 - 3i$(3+2i)+(1−5i)=(3+1)+(2+(−5))i=4−3i

Worked Example 2: Compute $(7 - i) - (4 + 3i)$(7−i)−(4+3i).

$(7 - i) - (4 + 3i) = (7 - 4) + (-1 - 3)i = 3 - 4i$(7−i)−(4+3i)=(7−4)+(−1−3)i=3−4i

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Multiplication

To multiply complex numbers, expand using the distributive law (like multiplying brackets) and use the fact that $i^2 = -1$i2=−1:

$(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i$(a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+(ad+bc)i

You do not need to memorise the formula — just expand and simplify.

Worked Example 3: Compute $(2 + 3i)(4 - i)$(2+3i)(4−i).

$(2 + 3i)(4 - i) = 2(4) + 2(-i) + 3i(4) + 3i(-i)$(2+3i)(4−i)=2(4)+2(−i)+3i(4)+3i(−i) $= 8 - 2i + 12i - 3i^2$=8−2i+12i−3i2 $= 8 + 10i - 3(-1)$=8+10i−3(−1) $= 8 + 10i + 3$=8+10i+3 $= 11 + 10i$=11+10i

Worked Example 4: Compute $(1 + i)^2$(1+i)2.

$(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$(1+i)2=1+2i+i2=1+2i−1=2i

Exam Tip: A common error is to forget that $i^2 = -1$i2=−1 when expanding brackets. Always simplify $i^2$i2 terms.

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Multiplying by a Real Number

Multiplying a complex number by a real number $k$k scales both parts:

$k(a + bi) = ka + kbi$k(a+bi)=ka+kbi

Example: $3(2 - 4i) = 6 - 12i$3(2−4i)=6−12i

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Division (Realising the Denominator)

To divide complex numbers, we multiply both numerator and denominator by the complex conjugate of the denominator. The complex conjugate of $c + di$c+di is $c - di$c−di.

$\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}$c+dia+bi​=(c+di)(c−di)(a+bi)(c−di)​=c2+d2(a+bi)(c−di)​

The key fact is:

$(c + di)(c - di) = c^2 + d^2$(c+di)(c−di)=c2+d2

This is always a real number (and positive when $c + di \neq 0$c+di=0), which is why this process is called realising the denominator.

Worked Example 5: Compute $\frac{3 + i}{1 - 2i}$1−2i3+i​.

Multiply numerator and denominator by $1 + 2i$1+2i:

$\frac{(3 + i)(1 + 2i)}{(1 - 2i)(1 + 2i)} = \frac{3 + 6i + i + 2i^2}{1 + 4} = \frac{3 + 7i - 2}{5} = \frac{1 + 7i}{5} = \frac{1}{5} + \frac{7}{5}i$(1−2i)(1+2i)(3+i)(1+2i)​=1+43+6i+i+2i2​=53+7i−2​=51+7i​=51​+57​i

Worked Example 6: Express $\frac{5}{2 + i}$2+i5​ in the form $a + bi$a+bi.

$\frac{5}{2 + i} = \frac{5(2 - i)}{(2 + i)(2 - i)} = \frac{10 - 5i}{4 + 1} = \frac{10 - 5i}{5} = 2 - i$2+i5​=(2+i)(2−i)5(2−i)​=4+110−5i​=510−5i​=2−i

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