The number which contain iota (i) values are known as imaginary Numbers. In mathematics representation of Imaginary Number is given by symbol 'i'. Let's discuss how to solve imaginary numbers.

Some steps to solve the imaginary numbers are given below:

Step 1: To solve imaginary numbers first of all we take an expression that contain imaginary values.

Step 2: Then we have to multiply first term of first Set to all values of the next set and we apply same procedure for the second term of first set.

Let's see how to simplifying imaginary numbers.

Suppose we have an expression (4 – i) (6 + 7i).

To simplify this expression we have to follow the above steps which are:

Step 1: To solve this expression first of all we have to multiply first value of first set to all terms of the next set and same procedure is applied to the next term of first set.

= (4 – i) (6 + 7i), we can write it as using above step:

= (4 * 6) + (4 * 7i) - (6 * I) – 7i

= 24 + 28i – 6i – 7i

= 24 + 28i – 6i - 7 (-1);

= 24 + 22i + 7; This is how we can simplifying imaginary numbers.

Here one basic condition of imaginary number is given as:

= i = √-1, the value of 'I' is √-1. Here if we use iota in the equations then we can write the Square root of negative numbers. For example: (√ s (√ t)), we can also write this given value as√ st.

This is how we can solve imaginary numbers.

Some steps to solve the imaginary numbers are given below:

Step 1: To solve imaginary numbers first of all we take an expression that contain imaginary values.

Step 2: Then we have to multiply first term of first Set to all values of the next set and we apply same procedure for the second term of first set.

Let's see how to simplifying imaginary numbers.

Suppose we have an expression (4 – i) (6 + 7i).

To simplify this expression we have to follow the above steps which are:

Step 1: To solve this expression first of all we have to multiply first value of first set to all terms of the next set and same procedure is applied to the next term of first set.

= (4 – i) (6 + 7i), we can write it as using above step:

= (4 * 6) + (4 * 7i) - (6 * I) – 7i

^{2}; on further solving we get:= 24 + 28i – 6i – 7i

^{2}, we can also write -1 in place of i^{2};= 24 + 28i – 6i - 7 (-1);

= 24 + 22i + 7; This is how we can simplifying imaginary numbers.

Here one basic condition of imaginary number is given as:

= i = √-1, the value of 'I' is √-1. Here if we use iota in the equations then we can write the Square root of negative numbers. For example: (√ s (√ t)), we can also write this given value as√ st.

This is how we can solve imaginary numbers.