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Is $f(z)=z^n$ holomorphic?

I have tested a number of other functions using the Cauchy Riemann equations $u_x=v_y$, $v_x=-u_y$. However in the case of $f(z)=z^n$ I cannot think of a way to find the functions $u(x,y)$ and $v(x,y)$ without using a binomial expansion of $(x+iy)^n$.

Any help or pointers is appreciated.

edit - the problem requires the use of the Cauchy - Riemann equations and not the formal definition of complex differentiation.

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  • $\begingroup$ What definition of holomorphic function do you know? $\endgroup$
    – agha
    Feb 8, 2015 at 22:47
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    $\begingroup$ What's wrong with using the binomial theorem? $\endgroup$ Feb 8, 2015 at 22:47
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    $\begingroup$ The exercises the problems are from requires the use of Cauchy Riemann equations, so I have been staying away from the formal definition for now $\endgroup$
    – Modulus
    Feb 8, 2015 at 22:56
  • $\begingroup$ If you want to avoid using the Binomial Theorem, then you can use the Product Rule. Start with $ f: z \mapsto z $, which is holomorphic. Then $ g: z \mapsto z^{2} $ is the product of two holomorphic functions, and so by the Product Rule, it is holomorphic. This is essentially the start of an induction argument. $\endgroup$ Feb 9, 2015 at 3:35
  • $\begingroup$ @BerrickFillmore see also my answer below. $\endgroup$
    – Mårten W
    Feb 9, 2015 at 7:54

3 Answers 3

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The easiest way is just the definition

$$\lim_{z\to a}{z^n-a^n\over z-a}=\lim_{z\to a}(z^{n-1}+az^{n-2}+\ldots +a^{n-2}z+a^{n-1})=na^{n-1}$$

by definition, since $a$ was arbitrary, the derivative exists at all points, $a\in\Bbb C$.

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  • $\begingroup$ Are you assuming that n>1? If, for example, $n = -1$, then, the limit does not exit at the origin. So is $z^{-1}$ holomorphic? $\endgroup$ Apr 23, 2021 at 14:50
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HINT: Show that if $f=u+iv$ and $g=\hat{u}+i\hat{v}$ satisfy the Cauchy-Riemann equations, then so does $fg$. Use this together with the fact that $z^n$ is a finite product of holomorphic functions (since $z$ is holomorphic).

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Consider the function $f(x,y)=(x+iy)^n$. Then the real and imaginary parts of this function are $$ u(x,y)=\frac{1}{2}((x+iy)^n+(x-iy)^n)\\ v(x,y)=\frac{1}{2i}((x+iy)^n-(x-iy)^n)$$ so $$ \frac{\partial u}{\partial x}=\frac{n}{2}((x+iy)^{n-1}+(x-iy)^{n-1}) $$ whereas $$ \frac{\partial v}{\partial y}= \frac{n}{2i}(i(x+iy)^{n-1}+i(x-iy)^{n-1}) $$

Verify also the other Cauchy-Riemann equation.

Is this legitimate? Yes, of course. We're just considering functions $\mathbb{R}\to\mathbb{C}$ and derivatives are perfectly defined as usual, with the usual properties.


If you don't trust this (but you should), you can do a proof by induction. Denote by $u_n(x,y)$ and $v_n(x,y)$ the real and imaginary parts of $f(x,y)=(x+iy)^n$. Then \begin{align} u_n(x,y)+iv_n(x,y)&=(u_{n-1}(x,y)+iv_{n-1}(x,y))(x+iy)\\ &= (xu_{n-1}(x,y)-yv_{n-1}(x,y))+i(xv_{n-1}(x,y)+yu_{n-1}(x,y)) \end{align} So \begin{align} u_n(x,y)&=xu_{n-1}(x,y)-yv_{n-1}(x,y)\\ v_n(x,y)&=xv_{n-1}(x,y)+yu_{n-1}(x,y) \end{align} Compute the partial derivatives and apply the induction hypothesis.

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