complex number

Question

multiply:-

               (2√-3+3√-2) by (4√-3-5√-2)

Answer 1

Let's start by defining the imaginary number i as √-1.  Therefore √-1 = i, so √-2 = 2 . i etc. 

 

(2√-3+3√-2) . (4√-3-5√-2) = (2 . √3 . i + 3 . √2. i) . (4 . √3 . i - 5 . √2 . i)

 

Using the long multiplication technique, we multiply each term of the first expression with each term in the second and add them up

 

This yields the following expression

 

(2 . √3 . i) . (4 . √3 . i) + (2 . √3 . i) . (- 5 . √2 . i) + (3 . √2. i) . (4 . √3 . i) + (3 . √2. i) . (- 5 . √2 . i)

 

(2 . √3 . i) . (4 . √3 . i) = 2 . 4 . √3 . √3 . i . i = 8 . 3 . (-1) = -24

 

(2 . √3 . i) . (- 5 . √2 . i) = 2 . (-5) . √6 . (-1) = 10√6

 

(3 . √2. i) . (4 . √3 . i) = 3 . 4 . √2 . √3 . i . i = -12√6

 

(3 . √2. i) . (- 5 . √2 . i) = 3 . (-5) . √2 . √2 . i . i = 30

 

So the answer is -24  + 10√6 - 12√6 + 30 = 6 - 2√6

 

This is a real, irrational number (not a complex number)