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I have some trouble with the initial condition of this equation: $f(z) = u(x+iy)+iv(x+iy)$ where $u(x+iy)=x^2-y^2+x$ and initial condition $f(i)=-1+i$.

Can someone please help me how to come to solution with initial condition ?

My solution is:

$x^2-y^2+x + i(2xy+y+C)$ where C is constant. I know I have to somehow replace the constant with the initial condition but I really don't know how to do it. Maybe it is simple but I don't have a clue...

Thank you for your answers.

I was trying a solution which was suggested on this:

$f(z) = x^2 - y^2 + xy + i(-(x^2/2) + y^2/2 +2xy + C), f(1+i)=-2i$

and I came to solution: $Ci=-2i \implies C = -2$ but it is marked as incorrect, it has to be $C=-1$ am I doing something wrong ? I substituted $x=1,y=i$...

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Just calculate the value of your function for $x=0$ and $y=1$ which corresponds with $i=0+i\cdot 1$ and then compare it to $-1+i$ to find $C$. –  savick01 Jan 7 '12 at 21:11

1 Answer 1

For $z=x+iy$: $$f(z)=x^2-y^2+x + i(2xy+y+C)$$

which gives: $$f(i)=0^2-i^2+0+i(2(0)(i)+i+C)=iC$$

(Note: In the above we substitute $x=0,y=i$.)

Since $f(i)=-1+i$ it follows that: $$iC = -1+i \implies C = 1+i$$

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I edited first post, I tried your solution on another function. –  velocityIs Jan 8 '12 at 9:27
    
@velocityIs: In your working for your second example you claim $Ci=-2i$. Are you saying that $f(1+i)=Ci$? Is this really the case? –  matt Jan 9 '12 at 1:07

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