How do you perform the operation and write the result in standard form given #(2+3i)^2+(2-3i)^2#?
1 Answer
Nov 12, 2016
Explanation:
# (2+3i)^2+(2-3i)^2 = {2^2+2(2)(3i)+(3i)^2} + {2^2-2(2)(3i)+(3i)^2} #
# :. (2+3i)^2+(2-3i)^2 = {4+12i+9i^2} + {4-12i+9i^2} #
# :. (2+3i)^2+(2-3i)^2 = 8+18i^2 #
Now,
# :. (2+3i)^2+(2-3i)^2 = 8-18 #
# :. (2+3i)^2+(2-3i)^2 = -10 #
