Multiply numerator and denominator by conjugate of denominator (1 + 2i). Standard method.

Express (2 + 3i)/(1 - 2i) in a + ib Form — Complex Division

Question

Express $\dfrac{2 + 3i}{1 - 2i}$ in the standard form $a + ib$ .

Solution — Step by Step

The denominator is $1 - 2i$ , so its conjugate is $1 + 2i$ . We multiply both numerator and denominator by this conjugate — the key move that eliminates $i$ from the denominator.

\frac{2 + 3i}{1 - 2i} \times \frac{1 + 2i}{1 + 2i}

We haven’t changed the value (we’re multiplying by 1), but now we can simplify.

Numerator: $(2 + 3i)(1 + 2i)$

= 2(1) + 2(2i) + 3i(1) + 3i(2i)

= 2 + 4i + 3i + 6i^2

Since $i^2 = -1$ :

= 2 + 7i + 6(-1) = 2 - 6 + 7i = -4 + 7i

Denominator: $(1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5$

This is why the conjugate trick works — the denominator always becomes a real number.

\frac{-4 + 7i}{5} = -\frac{4}{5} + \frac{7}{5}i

So $a = -\dfrac{4}{5}$ and $b = \dfrac{7}{5}$ .

Answer: $\dfrac{2 + 3i}{1 - 2i} = -\dfrac{4}{5} + \dfrac{7}{5}i$

Why This Works

The trick here is the algebraic identity $(a - b)(a + b) = a^2 - b^2$ . When $b = 2i$ , we get $b^2 = 4i^2 = -4$ , which is real. So the product $(1 - 2i)(1 + 2i)$ is guaranteed to be a positive real number.

This is directly analogous to rationalising the denominator in surds — something you already know from Class 9. Just like we multiply by $(\sqrt{2} + 1)$ to remove $\sqrt{2}$ from the denominator, we multiply by the conjugate to remove $i$ .

Every complex division problem in JEE and CBSE boards follows this same structure. Once this is muscle memory, these questions become pure calculation — no thinking required, just execution.

Alternative Method — Using the Formula Directly

If you’re in exam mode and want to skip the expansion, use this direct formula:

\frac{a + ib}{c + id} = \frac{ac + bd}{c^2 + d^2} + \frac{bc - ad}{c^2 + d^2}i

Here, $a = 2$ , $b = 3$ , $c = 1$ , $d = -2$ .

\text{Real part} = \frac{(2)(1) + (3)(-2)}{1^2 + (-2)^2} = \frac{2 - 6}{1 + 4} = \frac{-4}{5}

\text{Imaginary part} = \frac{(3)(1) - (2)(-2)}{1^2 + (-2)^2} = \frac{3 + 4}{5} = \frac{7}{5}

Memorise this formula for 1-mark MCQs in JEE Main. It saves 40-50 seconds per question. Just be careful with the signs on $d$ — students drop the negative sign more often than not.

Common Mistake

The most common error: forgetting that $i^2 = -1$ mid-expansion, or treating $6i^2$ as $+6$ instead of $-6$ . In the numerator expansion, $(2 + 3i)(1 + 2i)$ gives a $6i^2$ term — students in a hurry often write $-4 + 7i$ correctly but then flip the sign on $6i^2$ somewhere and get $+8 + 7i$ instead. Always write out $i^2 = -1$ explicitly until it becomes second nature.

A second trap: multiplying only the numerator by the conjugate and forgetting the denominator. You must multiply by $\dfrac{1+2i}{1+2i}$ , not just $1+2i$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Formula Directly

Common Mistake

Try These Next