equivalent to (4 + 6i)2?

Question

How would i do this?

Answer 1

As Parviz pointed out the problem is most likely written as (4 + 6i)², in which case you must solve it in the following way by expanding (4 + 6i)(4 + 6i)

 

(4 + 6i)(4 + 6i)

(4*4) + (4*6i) + (4*6i) + (6i*6i)

16 + 24i + 24i - 36

48i - 20

 

Make sure that you remember how to multiply two imaginary numbers, i*i=-1.

 

Let me know if you have any questions about this explanation.

Answer 2

( 4 + 6i)² = 16 +48i - 36 = - 20 + 48i

 

   The problem is to raise Complex number to the power of 2:

 

   ( 4 + 6i) ² = (4 +6i)(4 + 6i)

                       ( 4 + 6i) 4 + 6i( 4+ 6i)=

                        16 + 24i + 24i -36 = -16 + 48i

 

    Student should know by heart that a binomial raised to the power of 2, 

    

     results   in ( first term raised to 2nd power+ twice first term times 2nd term +

                       2nd term raised to 2nd power).

      That is how I wrote the result straight.

Answer 3

is it (4 + 6i) ^ 2 or (4 + 6i)(2)