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 ,

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

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



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.

  • $\begingroup$ Did you tried Cauchy-Riemann's equations? $\endgroup$ – k1.M May 12 '15 at 9:38

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.

  • $\begingroup$ 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

$f^{'}(z) = 2y + iv(x,y)$

$u_x=v_y$ 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) $b(x,y)$


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

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


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

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

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

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

Hence the answer...

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

  • $\begingroup$ That's some really long effort . +1 ^^ $\endgroup$ – Mann May 12 '15 at 17:50

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$ $x\to z$


Integrating this we will get,


Now using the given condition ,




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

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




  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.