Question

(2 + i)(3 - i)(1 + 2i)(1 - i)(3 + i)

Answer 1

Answer:

50+50i

Step-by-step explanation:

Given the expression (2 + i)(3 - i)(1 + 2i)(1 - i)(3 + i), we are to take the product of all the complex values. We must note that i² = -1.

Rearranging the expression [(3 - i)(3 + i)] [(2 + i)(1 - i)](1 + 2i)

On expansion

(3 - i)(3 + i)

=  9+3i-3i-i²

= 9-(-1)

= 9+1

(3 - i)(3 + i) = 10

For the expression (2 + i)(1 - i), we have;

(2 + i)(1 - i)

= 2-2i+i-i²

= 2-i+1

= 3-i

Multiplying 3-i with the last expression (1 + 2i)

(2 + i)(1 - i)(1 + 2i)

= (3-i)(1+2i)

= 3+6i-i-2i²

= 3+5i-2(-1)

= 3+5i+2

= 5+5i

Finally,  [(3 - i)(3 + i)] [(2 + i)(1 - i)(1 + 2i)]

= 10(5+5i)

= 50+50i

Hence,  (3 - i)(3 + i)(2 + i)(1 - i)(1 + 2i) is equivalent to 50+50i