Question

find the value of x²+y² if x +iy=(1+i)(2+i)(3+i)​

Answer 1

Solution :

Now, x + iy = (1 + i) (2 + i) (3 + i)

= (2 + i + 2i + i²) (3 + i)

= (2 + 3i + i²) (3 + i)

= (2 + 3i - 1) (3 + i) , since i² = - 1

= (1 + 3i) (3 + i)

= 3 + i + 9i + 3i²

= 3 + 10i + 3i²

= 3 + 10i - 3

= 10i

↣ x + iy = 10i

Comparing among both sides real and imaginary parts, we get

x = 0 and y = 10

Thus, x² + y²

= 0² + 10²

= 0 + 100

= 100

Answer 2

Answer: x² + y² = 100

Step-by-step explanation:

x+iy = (1+i)(2+i)(3+i)

       = [2+i+2i+i²](3+i)

       = (2+3i-1)(3+i)

       = (1+3i)(3+i)

       = (3+i+9i+3i²)

       = (3 + 10i -3)

x+iy  = 0 + 10i

By comparing imaginary and real parts

x = 0

y = 10

x² + y² = 0 + 10²

x² + y² = 100