1. ## Equation, Complex numbers

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

2. ## Re: Equation, Complex numbers

Originally Posted by kreshnik
$\displaystyle$(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$Here is a hint: If$\displaystyle a+bi=-1+2i$then$\displaystyle a=-1~\&~b=2$. 3. ## Re: Equation, Complex numbers I never wrote this: Originally Posted by plato here is a hint: If$\displaystyle a+bi=-1+2i$then$\displaystyle a=-1~\&~b=2$. you meant: Originally Posted by kreshnik$\displaystyle 3(2i-1)=a+bi$or Originally Posted by kreshnik$\displaystyle 6i-3=a+bi$so:$\displaystyle a+bi=-3+6i$hence:$\displaystyle a=-3\;\;\;,\;\;\;b=6$Am I wrong?$\displaystyle a+b=-3+6=3$or$\displaystyle a+b=-1+2=1$which one?, sorry for taking so long 4. ## Re: Equation, Complex numbers Originally Posted by kreshnik$\displaystyle a+b=-3+6=3\$
This is correct. The're no two options! Where does your other 'solution' come from?

5. ## Re: Equation, Complex numbers

Originally Posted by Siron
This is correct. The're no two options! Where does your other 'solution' come from?
I think Plato somehow didn't see number 3. But it's ok, now I know how to do It, thanks everyone.