Division of complex numbers is also a complex number.

In other words, the division of two complex numbers can be expressed in the standard form A + iB where A and B are real.

Division of a complex number z\(_{1}\) = p + iq by z\(_{2}\) = r + is ≠ 0 is defined as

\(\frac{z_{1}}{z_{2}}\) = \(\frac{pr + qs}{\sqrt{r^{2} + s^{2}}}\) + i\(\frac{qr - ps}{\sqrt{r^{2} + s^{2}}}\)

**Proof:**

Given z\(_{1}\) = p + iq by z\(_{2}\) = r + is ≠ 0

\(\frac{z_{1}}{z_{2}}\) = z1 ∙ \(\frac{1}{z_{2}}\) = z\(_{1}\) ∙ z\(_{2}\)\(^{-1}\) = (p + iq) . \(\frac{r - is}{\sqrt{r^{2} + s^{2}}}\) = \(\frac{pr + qs}{\sqrt{r^{2} + s^{2}}}\) + i\(\frac{qr - ps}{\sqrt{r^{2} + s^{2}}}\)

Again,

\(\frac{z_{1}}{z_{2}}\) = \(\frac{p + iq}{r + is}\) = \(\frac{p + iq}{r + is}\) × \(\frac{r - is}{r - is}\) = \(\frac{(pr + qs) + i(qr - ps)}{\sqrt{r^{2} + s^{2}}}\) = A + iB where A = \(\frac{pr + qs}{\sqrt{r^{2} + s^{2}}}\) and B = \(\frac{qr - ps}{\sqrt{r^{2} + s^{2}}}\) are real.

Therefore, quotient of two complex numbers is a complex number.

For example, if z\(_{1}\) = 2 + 3i and z\(_{2}\) = 4 - 5i, then

\(\frac{z_{1}}{z_{2}}\) = \(\frac{2 + 3i}{4 - 5i}\) = \(\frac{2 + 3i}{4 - 5i}\) × \(\frac{4 + 5i}{4 + 5i}\) = \(\frac{(2 × 4 - 3 × 5) + (2 × 5 + 3 × 4)i}{4^{2} - 5^{2} × i^{2}}\)

= \(\frac{(8 - 15) + (10 + 12)i}{16 + 25}\)

= \(\frac{-7 + 22i}{41}\)

= \(\frac{-7}{41}\) + \(\frac{22}{41}\)i

Solved example on division of two complex numbers:

Find the quotient when the complex number 5 + √2i divided by the complex number 1 - √2i.

**Solution:**

\(\frac{5 + √2i}{1 - √2i}\)

= \(\frac{5 + √2i}{1 - √2i}\) × \(\frac{1 + √2i}{1 + √2i}\)

= \(\frac{5 + 5√2i + √2i + 2i^{2}}{1^{2} – (√2i)^{2}}\)

= \(\frac{5 + 6√2i - 2}{1 - 2(-1)}\)

= \(\frac{3 + 6√2i}{3}\)

= 1 + 2√2i

**11 and 12 Grade Math****From Division of Complex Numbers** **to HOME PAGE**

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.