# 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 #