# Division of a Complex Number

To divide a complex number means rationalizing the denominator of the complex number. It is done by multiplying the numerator and the denominator by the complex conjugate of the denominator. This will eliminate the imaginary part, giving a real number at the end.

Let us divide a complex number z1 = a1 + ib1 by another complex number z2 = a2 + ib2.

Then, on dividing z1 and z2, the quotient  obtained is: ${\dfrac{z_{1}}{z_{2}}=\dfrac{a_{1}+ib_{1}}{a_{2}+ib_{2}}}$.

## Formula

The formula to calculate the quotient is given below:

${\dfrac{a_{1}a_{2}+b_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}+\left( \dfrac{a_{2}b_{1}-a_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}\right) i}$

Let us now divide z1 = 3 + 5i by z2 = 1 – 6i using the above formula.

Thus, ${\dfrac{3+5i}{1-6i}}$

First, we will find the complex conjugate of the denominator.

Here, the complex conjugate of 1 – 6i is 1 + 6i.

Now, on eliminating the imaginary part in the denominator by multiplying its conjugate with the numerator and denominator, we get

${\dfrac{3+5i}{1-6i}\times \dfrac{1+6i}{1+6i}}$

Applying the identity (a + b)(a – b) = a2 – b2 in the denominator and substituting the value i2 = -1, we get,

${\dfrac{\left( 3+5i\right) \left( 1+6i\right) }{1^{2}-\left( 6i\right) ^{2}}}$

Simplifying further, we get ${\dfrac{3+5i+18i+30i^{2}}{1+36}}$

= ${\dfrac{3+23i-30}{37}}$

= ${\dfrac{-27+23i}{37}}$

= ${\dfrac{-27}{37}+\dfrac{23i}{37}}$

Thus, ${\dfrac{3+5i}{1-6i}}$ = ${\dfrac{-27}{37}+\dfrac{23i}{37}}$.

Similarly, we can divide the complex numbers in the polar form.

## In the Polar Form

The division of complex numbers z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2) in polar form is given by the formula:

${\dfrac{z_{1}}{z_{2}}=\dfrac{r_{1}\left( \cos \theta _{1}+i\sin \theta _{1}\right) }{r_{2}\left( \cos \theta _{2}+i\sin \theta _{2}\right) }}$

⇒ ${\dfrac{z_{1}}{z_{2}}}$ = r(cosθ + isinθ), here r = ${\dfrac{r_{1}}{r_{2}}}$, and θ = (θ1 – θ2).

Now. let us divide the complex numbers z1 = 7(cos190° + isin190°) by z2 = 2(cos90° + isin90°) using the above formula.

${\dfrac{z_{1}}{z_{2}}=\dfrac{7\left( \cos 190^{\circ }+i\sin 190^{\circ }\right) }{2\left( \cos 90^{\circ }+i\sin 90^{\circ }\right) }}$

Here, the complex conjugate of cos90° + isin90° is cos90° – isin90°.

On eliminating the imaginary part in the denominator by multiplying the complex conjugate with the numerator and denominator, we get,

${\dfrac{7\left( \cos 190^{\circ }+i\sin 190^{\circ }\right) }{2\left( \cos 90^{\circ }+i\sin 90^{\circ }\right) }\times \dfrac{\left( \cos 90^{\circ }-i\sin 90^{\circ }\right) }{\left( \cos 90^{\circ }-i\sin 90^{\circ }\right) }}$

Applying the identity (a + b)(a – b) = a2 – b2 in the denominator and substituting the value i2 = -1, we get ${\dfrac{7\left( \cos 190^{\circ }+i\sin 190^{\circ }\right) \left( \cos 90^{\circ }-i\sin 90^{\circ }\right) }{2\left( \cos ^{2}90^{\circ }-i^{2}\sin ^{2}90^{\circ }\right) }}$

= ${\dfrac{7\left( \cos 190^{\circ }+i\sin 190^{\circ }\right) \left( \cos 90^{\circ }-i\sin 90^{\circ }\right) }{2\left( \cos ^{2}90^{\circ }+\sin ^{2}90^{\circ }\right) }}$

Simplifying further, we get ${\dfrac{7}{2}\left( \cos 190^{\circ }+i\sin 190^{\circ }\right) \left( \cos 90^{\circ }-i\sin 90^{\circ }\right)}$

= ${\dfrac{7}{2}\left[ \cos \left( 190^{\circ }-90^{\circ }\right) +i\sin \left( 190^{\circ }-90^{\circ }\right) \right]}$

= ${\dfrac{7}{2}\left( \cos 100^{\circ }+i\sin 100^{\circ }\right)}$

= r (cosθ + sinθ), where r = ${\dfrac{7}{2}}$ and θ = (190° – 90°) = 100°.

## Solved Examples

Solve: ${\dfrac{5-\sqrt{2}i}{-1-\sqrt{2}i}}$

Solution:

Here, ${\dfrac{5-\sqrt{2}i}{-1-\sqrt{2}i}}$
= ${\dfrac{5-\sqrt{2}I}{-1-\sqrt{2}i}\times \dfrac{-1+\sqrt{2}i}{-1+\sqrt{2}i}}$
= ${\dfrac{\left( 5-\sqrt{2}i\right) \left( -1+\sqrt{2}i\right) }{\left( -1\right) ^{2}-\left( \sqrt{2}\right) ^{2}i^{2}}}$
= ${\dfrac{-5+\sqrt{2}i+5\sqrt{2}i-2i^{2}}{1+2}}$
= ${\dfrac{-5+6\sqrt{2}i+2}{3}}$
= ${\dfrac{-3+6\sqrt{2}i}{3}}$
= ${-1+2\sqrt{2}i}$

Find the quotient by dividing the complex numbers 7 – 5i with 8 – 2i.

Solution:

Here, ${\dfrac{7-5i}{8-2i}}$
= ${\dfrac{7-5i}{8-2i}\times \dfrac{8+2i}{8+2i}}$
= ${\dfrac{\left( 7-5i\right) \left( 8+2i\right) }{8^{2}-\left( 2i\right) ^{2}}}$
= ${\dfrac{56-40i+14i-10i^{2}}{64+4}}$
= ${\dfrac{56-26i+10}{68}}$
= ${\dfrac{66-26i}{68}}$
= ${\dfrac{66}{68}-\dfrac{26i}{68}}$
= ${\dfrac{33}{34}-\dfrac{13i}{34}}$
Thus, ${\dfrac{7-5i}{8-2i}}$ = ${\dfrac{33}{34}-\dfrac{13i}{34}}$.