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Mathematics · Ch 4 — Complex Numbers and Quadratic Equations

Division of Two Complex Numbers

4.3.4

Division of Two Complex Numbers

The Meaning of Division

Division of complex numbers is defined in a way that keeps the result a complex number. For any two complex numbers z1z_1 and z2z_2, with z2≠0z_2 \neq 0, the quotient z1z2\frac{z_1}{z_2} is defined as:

z1z2=z1⋅1z2\frac{z_1}{z_2} = z_1 \cdot \frac{1}{z_2}

That is, dividing z1z_1 by z2z_2 means multiplying z1z_1 by the multiplicative inverse of z2z_2. This definition is natural — it mirrors the real-number idea that division is multiplication by the reciprocal.

Watch out

The condition z2≠0z_2 \neq 0 is essential. Division by the complex number 0+0i0 + 0i is undefined, just as division by zero is undefined in real numbers.

The Method: Rationalising the Denominator

In practice, we never directly compute 1z2\frac{1}{z_2} as a separate step. Instead, we use a technique that eliminates the imaginary part from the denominator — the same rationalisation trick you use for surds.

Given z1=a+ibz_1 = a + ib and z2=c+idz_2 = c + id (with c+id≠0c+id \neq 0), we multiply numerator and denominator by the complex conjugate of the denominator:

z1z2=a+ibc+id=(a+ib)(c−id)(c+id)(c−id)\frac{z_1}{z_2} = \frac{a+ib}{c+id} = \frac{(a+ib)(c-id)}{(c+id)(c-id)}

The denominator becomes a real number: (c+id)(c−id)=c2+d2(c+id)(c-id) = c^2 + d^2. So:

a+ibc+id=(a+ib)(c−id)c2+d2\frac{a+ib}{c+id} = \frac{(a+ib)(c-id)}{c^2 + d^2}

Now expand the numerator, separate real and imaginary parts, and you have the quotient in standard form x+iyx + iy.

Tip

Always check that c2+d2≠0c^2 + d^2 \neq 0 — but since z2≠0z_2 \neq 0, at least one of cc or dd is non-zero, so c2+d2>0c^2 + d^2 > 0. The denominator is always a positive real number.

Worked Example from the Textbook

Let z1=6+3iz_1 = 6 + 3i and z2=2−iz_2 = 2 - i. Find z1z2\frac{z_1}{z_2}.

Step 1: Write the quotient as multiplication by the inverse:

6+3i2−i=(6+3i)⋅12−i\frac{6+3i}{2-i} = (6+3i) \cdot \frac{1}{2-i}

Step 2: Find 12−i\frac{1}{2-i} by rationalising:

12−i=12−i⋅2+i2+i=2+i(2−i)(2+i)=2+i22+(−1)2=2+i4+1=2+i5\frac{1}{2-i} = \frac{1}{2-i} \cdot \frac{2+i}{2+i} = \frac{2+i}{(2-i)(2+i)} = \frac{2+i}{2^2 + (-1)^2} = \frac{2+i}{4+1} = \frac{2+i}{5}

Step 3: Multiply:

6+3i2−i=(6+3i)⋅2+i5=15(6+3i)(2+i)\frac{6+3i}{2-i} = (6+3i) \cdot \frac{2+i}{5} = \frac{1}{5} (6+3i)(2+i)

Step 4: Expand the product:

(6+3i)(2+i)=6⋅2+6⋅i+3i⋅2+3i⋅i=12+6i+6i+3i2(6+3i)(2+i) = 6\cdot 2 + 6\cdot i + 3i\cdot 2 + 3i\cdot i = 12 + 6i + 6i + 3i^2

Since i2=−1i^2 = -1:

=12+12i+3(−1)=12+12i−3=9+12i= 12 + 12i + 3(-1) = 12 + 12i - 3 = 9 + 12i

Step 5: Divide by 5:

6+3i2−i=9+12i5=95+125i\frac{6+3i}{2-i} = \frac{9 + 12i}{5} = \frac{9}{5} + \frac{12}{5}i

Note

The textbook shows a slightly different path — it first writes 12−i\frac{1}{2-i} in the form c+idc2+d2\frac{c+id}{c^2+d^2} using the formula 1c+id=c−idc2+d2\frac{1}{c+id} = \frac{c-id}{c^2+d^2}, then multiplies. Both methods are equivalent; the rationalisation approach is usually faster.

The General Formula for the Quotient

From the method above, we can write a direct formula. For z1=a+ibz_1 = a+ib and z2=c+idz_2 = c+id (c+id≠0c+id \neq 0):

a+ibc+id=ac+bdc2+d2+i⋅bc−adc2+d2\frac{a+ib}{c+id} = \frac{ac + bd}{c^2 + d^2} + i \cdot \frac{bc - ad}{c^2 + d^2}

This formula is worth knowing, but in practice it's safer to remember the rationalisation procedure — it's less error-prone.

Important

The quotient of two complex numbers is always a complex number. The denominator c2+d2c^2+d^2 is a positive real number, so the result is guaranteed to be of the form x+iyx + iy with x,y∈Rx, y \in \mathbb{R}.

Key Properties of Division

The textbook does not list separate numbered properties for division alone — division inherits its properties from multiplication by the inverse. However, two results are central:

Property 1 (Uniqueness of the quotient): For given z1z_1 and z2≠0z_2 \neq 0, the quotient z1z2\frac{z_1}{z_2} is unique. This follows because the multiplicative inverse of z2z_2 is unique, and multiplication by a fixed complex number is a well-defined operation.

Property 2 (Division by a real number): If z2=kz_2 = k is a real number (k≠0k \neq 0), then:

a+ibk=ak+ibk\frac{a+ib}{k} = \frac{a}{k} + i\frac{b}{k}

This is just the special case d=0d=0 in the general formula — the denominator c2+d2c^2+d^2 becomes k2k^2, and the numerator simplifies.

›Proof

Proof of Property 2:

Let z1=a+ibz_1 = a+ib and z2=kz_2 = k (real, k≠0k \neq 0). Then:

a+ibk=(a+ib)⋅1k=ak+ibk\frac{a+ib}{k} = (a+ib) \cdot \frac{1}{k} = \frac{a}{k} + i\frac{b}{k}

This is immediate from the definition of division as multiplication by the reciprocal, and the fact that the reciprocal of a real number kk is 1k\frac{1}{k} (a real number).

Common Pitfall to Avoid

A frequent mistake is to treat division of complex numbers like division of real numbers — trying to "separate" the denominator term-by-term. For example:

a+ibc+id≠ac+ibd\frac{a+ib}{c+id} \neq \frac{a}{c} + i\frac{b}{d} …