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please have a look at the question, Necessary condition for analyticity of $f(x+iy)=x^3+ax^2y+bxy^2+cy^3$ to solve the question i started from
f(x+iy)=u(x,y)+iv(x,y) since it is analytic it will hold the C-R equations.
$$ \begin{cases} u_x = 3x^2+2axy+by^2\\ u_y = ax^2+2bxy+3cy^2 \end{cases} $$
$$ \begin{cases} v_x(x,y) &= 0 \\ v_y(x,y) &= 0 \end{cases} $$
so, $$ \begin{cases} 3x^2+2axy+by^2&=0\\ ax^2+2bxy+3cy^2&=0 \end{cases} $$
ux,uy are partial derivative of u wrt x,y resp. Similarly applies for v.
After equating the equation 1 and 2, the answers comes out to be a=3,b=3,c=1
the answer seems to differ from the one that is in the given above link.

Is there some error in the answer?

kiv
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  • You are assuming that $a,b,c$ are real numbers. They are not given to be and if they are real then the function is never analytic. – Kavi Rama Murthy Apr 03 '20 at 09:56
  • @KaviRamaMurthy well after equating the equations I am getting values of a,b,c in real numbers. – kiv Apr 03 '20 at 10:07
  • Your formulas for $u$ and $v$ are wrong. It is not true that $v=0$. – Kavi Rama Murthy Apr 03 '20 at 10:08
  • For $y=0$, this gives to you that $f(x)=x^3$ Hence your function is equal to the analytic function $g(z)=z^3$ on $\mathbb{R}$. Hence if $f$ is analytic you have $f(z)=z^3=(x+iy)^3$. – Kelenner Apr 03 '20 at 12:21

1 Answers1

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You have to write down the real and imaginary parts of $a,b$ and $c$ and then the real and imaginary parts of $f$ to find out what $u$ and $v$ are.

Kavi Rama Murthy
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