Hello.

Is given:

$\displaystyle $(1+i)^2-(2-\text{i})^2=a+bi\;\;\;\text{find: a+b}$

$\displaystyle $A)\;1$

$\displaystyle $B)\;2$

$\displaystyle $C)\;3$

$\displaystyle $D)\;4$

I have tried:

$\displaystyle $(1+i)^2-(2-i)^2=a+bi$

$\displaystyle $1^2+2*1*\text{i}+\text{i}^2-(2^2-2*2*\text{i}+\text{i}^2)=a+bi$

$\displaystyle $2i-4+4i+1=a+bi$

$\displaystyle 6i-3=a+bi$

$\displaystyle 3(2i-1)=a+bi$

Here I'm stuck... Any help would be great. Thanks