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Calculate the derivative of $f(z)=\frac{(1+z^2)^4}{z^2}$

I know that this function is discontinous at $z=0$, what I did is just calculate the derivative the same manner as is done with real functions. $$\begin{align}f'(z) &=\frac{4(1+z^2)^32z*z^2-(1+z^2)^42z}{z^4}\\ &=\frac{8z^3(1+z^2)^3-(1+z^2)^42z}{z^4}\\ &=\frac{(1+z^2)^3[8z^3-2z(1+z^2)]}{z^4}\\ &=\frac{(1+z^2)^3[8z^3-2z-2z^3]}{z^4}\\ &=\frac{(1+z^2)^3(6z^3-2z)}{z^4}\end{align}$$

But I have some doubts: 1)I always need to check the Cauchy-Riemann conditions? 2)Mathematically speaking it is wrong to calculate the derivative that way?

The second question is why it had to write the function in the form $f(z)=u+iv$ with $u(x,y)$ and $iv(x,y)$ and make the derivative using $f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}$ can be very laborious.

Cameron Buie
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Roland
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1 Answers1

1

You can essentially prove like in the real case the usual formulas for derivatives. So

  • the derivative is linear: $(\alpha f(z)+\beta g(z))'=\alpha f'(z)+\beta g'(z)$

  • the product rule: $(f(z)g(z))'=f'(z)g(z)+f(z)g'(z)$

  • the quotient rule: $(f(z)/g(z))'=(f'(z)g(z)-f(z)g'(z))/(g(z)^2)$, where $g(z)\ne0$

  • polynomials: $(z^n)'=nz^{n-1}$

  • the chain rule: $f(g(z))'=f'(g(z))g'(z)$

Since existence of the derivative in an open set is sufficient for a function to be holomorphic in that open set, it follows that quotients of holomorphic functions are holomorphic in the set where the denominator doesn't vanish (which can't have limit points).

Thus you're surely allowed to compute the derivative as you did.

egreg
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