Question

Find the product of (x + (3+5i))^2

Answer 1

x² + 6x + 10xi -16 + 30i

Explanation:

(x + (3 + 5i) )²

use the binomial theorem:

(a + b)² = a² + 2ab + b² to expand (x + (3 + 5i) )²

substitute the values,

x² + (6 + 10i)x + (-16 + 30i)

therefore

x² + 6x + 10xi -16 + 30i

Answer 2

x^2 +6x +10xi -16 +30i

Explanation:

(x + (3+5i))^2

Rewriting as

(x + (3+5i)) (x + (3+5i))

Let 3+5i = m

(x + m) (x + m)

FOILing

x^2 + xm+xm+m^2

x^2 +2xm + m^2

Replacing m with 3+5i

x^2 +2x (3+5i) +(3+5i)^2

Distribute

x^2 +6x +10xi +(3+5i)^2

Now FOIL (3+5i)^2

( 3+5i) ( 3+5i)

3*3+ 3*5i + 3*5i+ 25i^2

= 9 +15i+15i +25(-1)

9+30i -25

-16 +30 i

Replacing that in the equation

x^2 +6x +10xi -16 +30i