# Division of Complex Numbers

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**Division of complex numbers :**We know that for any complex numbers z

_{1}and z

_{2}(≠0), there exists complex number z such that

z

_{1}= z . z

_{2}

This number z is denoted by $\frac{z1}{z2}$ called the division of complex numbers z

_{1}by z

_{2}(≠ 0). In fact, division of complex-numbers is the inverse operation of multiplication.

$\frac{z1}{z2}=\frac{(a+ib)}{(c+id)}=\frac{(a+ib)(c-id)}{(c+id)(c-id)}$

= $\frac{(ac+bd)}{c^{2} + d^{2}}+ \frac{(bc - ad)i}{c^{2} + d^{2}}$

## Properties of division of Complex Numbers

**Closure property:**The division of two complex-numbers is , by definition , a complex number. Hence, the set of complex numbers is closed under division.

Commutative and associative properties are not true for division of complex-numbers.

**Examples on division of complex-numbers**

**Divide the following :**

1) (8 + i) ÷ (5-i)

**Solution :**$\frac{(8+i)}{(5-i)} $

( multiply numerator and denominator by the conjugate of denominator)

= $\frac{(8+i)}{(5-i)}\frac{(5+i)}{(5+i)}$

=$\frac{(40 +8i + 5i +i^{2})}{(25-i^{2})}$

=$\frac{(40 +13i -1)}{(25+1)}$

= $\frac{(39+ 13i)}{(26)}$

= $\frac{39}{26}+\frac{13i}{26}$

= $\frac{3}{2}+\frac{i}{2}$

2) Write the complex z = $\frac{(2 +i)}{(1+i)(1-2i))}$

**Solution :**We have, $\frac{(2 +i)}{(1+i)(1-2i))}$

=$\frac{(2 +i)}{(1 -2i +i -2i^{2})}$

= $\frac{(2 +i)}{(1 -i +2)}$

=$\frac{(2 +i)}{(3 - i )}$

= $\frac{(2 +i)(3+i)}{(3-i)(3 +i)}$

= $\frac{(6 +2i + 3i +i^{2})}{(9-i^{2})}$

=$\frac{(6 +5i -1)}{(9+1)}$

=$\frac{(5 +5i)}{(10)}$

=$\frac{1}{2}+ \frac{1}{2}i$ = x + iy

∴ x =$\frac{1}{2}$ and y = $\frac{1}{2}$

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