1. ## Equation, Complex numbers

Hello.

Is given:

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

$A)\;1$

$B)\;2$

$C)\;3$

$D)\;4$

I have tried:

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

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

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

$6i-3=a+bi$

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

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

2. ## Re: Equation, Complex numbers

Originally Posted by kreshnik
$(1+i)^2-(2-\text{i})^2=a+bi\;\;\;\text{find: a+b}$
$A)\;1$
$B)\;2$
$C)\;3$
$D)\;4$
Here is a hint:
If $a+bi=-1+2i$ then $a=-1~\&~b=2$.

3. ## Re: Equation, Complex numbers

I never wrote this:
Originally Posted by plato
here is a hint:
If $a+bi=-1+2i$ then $a=-1~\&~b=2$.
you meant:

Originally Posted by kreshnik
$3(2i-1)=a+bi$
or
Originally Posted by kreshnik
$6i-3=a+bi$
so: $a+bi=-3+6i$

hence: $a=-3\;\;\;,\;\;\;b=6$

Am I wrong?

$a+b=-3+6=3$
or
$a+b=-1+2=1$

which one?, sorry for taking so long

4. ## Re: Equation, Complex numbers

Originally Posted by kreshnik

$a+b=-3+6=3$
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.