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An analytic function $f(z)$ is such that $\Re\{f'(z)\} =2y$ and $f(1+i)=2$. Then the imaginary part of $f(z)$ is

  1. $-2xy$

  2. $x^2-y^2$

  3. $2xy$

  4. $y^2-x^2$

Here, by using Milne Thomson's method I get $f'(z)=\int{-2i}$ $dz$ $=-2iz+c$, $c$ is the ingrating constant.

Now again integrating I got $f(z) =-iz^2+cz+d$, $d$ is another integrating constant.

But by only one condition I can not find out the value of two constants $c$ and $d$. But if I omit $c$, then using the given condition I get $\Im(f(z))$ as $y^2-x^2$.

I'm really confused and cannot understand what to do actually. Should I omit $c$ in the first integration?

Is there any other method for finding the solution?

Update:(21st May-2015)

Using Cauchy-Riemann equations we get $f'(z)=2y-2ix$. Then ,

$$f(z)=\int (2y-x)(\,dx+i\,dy)+C$$

$$=\int 2y\,dx+2i\int y\,dy-2i\int x\,dx+2\int x\,dy+C$$

$$=2xy+iy^2-ix^2+2xy+C$$

$$=4xy+i(y^2-x^2)+C.$$

So, if we neglect the constant or take it as a real constant then we get the imaginary part is $y^2-x^2$.

But I have a confusion about the integration. Is the integration correct ? If wrong then please tell me why it is incorrect ?

There is no difficulty about the answer of Daniel Fischer. See my update only.

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  • $\begingroup$ Did you tried Cauchy-Riemann's equations? $\endgroup$ – k1.M May 12 '15 at 9:38
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You are right, the data aren't sufficient to uniquely determine $f$.

Since we are given $\operatorname{Re} f'(z)$, we know that $c$ is purely imaginary, but it can be any purely imaginary constant. Before we take $f(1+i) = 2$ into account, we write the candidate for $f$ in the form

$$f(z) = -iz^2 + c(z - 1-i) + \tilde{d},$$

at which point $f(1+i) = 2$ reduces to $\tilde{d} = 0$, and we see that $\operatorname{Im} f(z)$ can be

$$y^2 - x^2 + r(x-1)$$

with an arbitrary $r\in\mathbb{R}$.

Among the four given choices, there is however only one that is possible.

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  • $\begingroup$ as you say $r \in \mathbb R$ is arbitrary, so I can put $r=0$ and get the required solution? $\endgroup$ – adember May 12 '15 at 9:48
  • $\begingroup$ It depends a bit on the situation. If it is a multiple choice test, there is only one possibility among the offered choices, so you pick that, i.e. choose $r = 0$. If you can give a longer answer, say that $\operatorname{Im} f(z)$ is not uniquely determined by the given constraints, and that all possibilities are of the form $y^2-x^2 + r(x-1)$ with $r\in \mathbb{R}$, where the choice $r = 0$ leads to option 4. $\endgroup$ – Daniel Fischer May 12 '15 at 9:55
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$f^{'}(z) = 2y + iv(x,y)$

$u_x=v_y$ and $v_x = - u_y$

$0 = v_y $ and $2 = -v_x$

$v_x = -2$

Integrating with respect to $'x'$

$v = -2x + \phi(y)$

differentiating with respect to $'y'$

$v_y = 0 + \phi^{'}(y)$

therefore $\phi^{'}(y) = v_y$

But we know that $v_y = 0$, hence $\phi^{'}(y) = 0$.

therefore $v = -2x + c$.

Hence $f'(z)=2y + i(-2x+c) $

Using c-r equations

$f'(z) = a_x + i b_x$

$a_x = 2y$ and $b_x = -2x + c$

we need to find $Im(f)$ (i.e) $b(x,y)$

Method1:

so, integrating with respect to $x$,

$b = -x^2 + xc+ \chi(y)$

Now, $f(x+iy) = a+ib$ ,$f(1+i) = 2 + i0$

so,$ 0 = -(1)^2 + (1)(c) + \chi(y)$

$\chi(y) = c+1 $

Substitute, $b= xc - x^2 + c+1$

hence $im(f) = xc - x^2 + c+1$.

method 2: instead of finding $b(x,y)$ find $a(x,y)$

now $a_x = 2y$

integrating with respect to $'x'$, we get

$a = 2xy + g(y); f(x+iy) = a + ib$ (i.e) $f(1+i) = 2+i0$

therefore when $x=1,y=1$ implies $a=2$ and $b=0$

$2 = 2(1)(1) + g(y)$

$g(y) = 0; a = 2xy;$

now $f(z) = 2xy + i b(x,y)$

$a_x = 2y$ and $a_y = 2x$

By C-R equations, $a_x = b_y$ and $b_x = - a_y$

$b_y =2y$; integrating

$b=y^2 + h(x)$

Differentiating

$b_x = h^{'}(x)$

$-2x = h^{'}(x)$

$h(x) = -x^{2}$

so $b = y^2 - x^2$..

Hence the answer...

If there is a mistake, please let me know it.

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  • $\begingroup$ That's some really long effort . +1 ^^ $\endgroup$ – Mann May 12 '15 at 17:50
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Let $f'(z)=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}$

Then, it is given that $\frac {\partial u}{\partial x}=2y$

Integrating this we get,

$u=2xy+\lambda(y)$ Where $\lambda (y)$ is an unknown function of $y$ that might have been lost in partial.

Calculating $\frac{\partial u}{\partial y}=2x+\lambda'(y)$

From CR equations, $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$

It all gives us $f'(z)=2y-i(2x+\lambda'(y))$

If $z=x+iy$ then as $y\to 0$$\implies$ $x\to z$

$f'(z)=-i(2z+\lambda'(0))$

Integrating this we will get,

$f(z)=-iz^2-iz\lambda'(0)+c_1$

Now using the given condition ,

$f(1+i)=2-i\lambda'(0)+\lambda'(0)+c_1$

$2=2-i\lambda'(0)+\lambda'(0)+c_1$

$0=c_1+\lambda'(0)-i\lambda'(0)$

Comparing real and imaginary part and assuming $c_1$ is real.

$\lambda'(0)=0$ and $c_1=0$

Gives:

$f(z)=iz^2$

$f(z)=2xy+i(y^2-x^2)$

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  • $\begingroup$ Why you assume that $C_1$ is real ? $\endgroup$ – Empty May 12 '15 at 15:44
  • $\begingroup$ That's the only assumption which gives answer compatible with options. $\endgroup$ – Mann May 12 '15 at 15:51

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