Question

How do you divide imaginary numbers?

Answer 1

Hint: We assume two complex numbers as a divisor and a dividend. We describe the concept of conjugate number for a complex number. Then we use the conjugate form to divide imaginary numbers. We find the reduced form of the division.

Complete step-by-step answer:
We are going to describe the concept of conjugate number for a complex number. Then we describe how to divide imaginary numbers using conjugate form.
For a complex number of forms $ z=x+iy $ , its conjugate complex number will be $ \overline{z}=x-iy $ .
The multiplication of $ z\overline{z} $ will be
 $ z\overline{z}=\left( x+iy \right)\left( x-iy \right) $ .
Simplifying we get $ z\overline{z}={{x}^{2}}+{{y}^{2}} $ . It is equal to the square of modulus value of $ z=x+iy $ .
Therefore, $ {{\left| z \right|}^{2}}=z\overline{z} $ .
In case of a division there has to be a divisor and a dividend. We assume those as complex numbers.
Let the dividend be $ a+ib $ and the divisor $ c+id $ .
Therefore, the form of the division is $ \dfrac{a+ib}{c+id} $ .
Now to solve the division we always need to take the conjugate form of the divisor and multiply with both numerator and denominator.
We are taking the conjugate form of the divisor $ c+id $ which is $ c-id $ and multiply both numerator and denominator of
 $ \dfrac{a+ib}{c+id} $ . We get $ \dfrac{\left( a+ib \right)\left( c-id \right)}{\left( c+id \right)\left( c-id \right)} $ . We complete the multiplication to get
 $ \dfrac{\left( a+ib \right)\left( c-id \right)}{\left( c+id \right)\left( c-id \right)}=\dfrac{\left( ac+bd \right)-i\left( ad-bc \right)}{{{c}^{2}}+{{d}^{2}}} $ .
This form will give the reduced form of $ \dfrac{\left( ac+bd \right)}{{{c}^{2}}+{{d}^{2}}}-i\dfrac{\left( ad-bc \right)}{{{c}^{2}}+{{d}^{2}}} $ .

Note: We have to remember that we don’t always need to follow process. Depending on the value of the complex number we can always change the multiplication process to ease off the multiplication process.