# Show that $F'(Z)=\frac{1}{Z}$ if $F(Z)=\log Z$ and $\frac{-\pi}{4}< \operatorname{arg}Z<\frac{7\pi}{4}$

Show that $F'(Z)=\frac{1}{Z}$ if $F(Z)=\log Z$ and $\frac{-\pi}{4}< \operatorname{arg}Z< \frac{7\pi}{4}$. Can I use definition of derivative here? What is the reason of adding condition on $\operatorname{arg}Z$? What will happen if the range of the argument is changed?

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"$-\frac{\pi}{4}\lt \mathrm{arg}Z\gt\frac{7\pi}{4}$" means that you want $\mathrm{arg}Z$ to be simultaneously greater than $-\frac{\pi}{4}$ and greater than $\frac{7\pi}{4}$. That's the same as just saying $\mathrm{arg}Z\gt \frac{7\pi}{4}$. Is that really what you meant, or di dyou mean $$-\frac{\pi}{4}\lt\mathrm{arg}(Z)\lt \frac{7\pi}{4}\ ?$$ –  Arturo Magidin Mar 25 '12 at 5:39
@ArturoMagidin: Thank you for you comment. I mean $-\frac{\pi}{4}<arg(Z)<\frac{7\pi}{4}$. I have fix it. –  Jite Mar 25 '12 at 5:48
\mathrm{arg} Z doesn't look the same as \operatorname{arg} Z. The latter has proper spacing before and after "arg". (I changed it.) –  Michael Hardy Mar 25 '12 at 15:00

In the first place you should use your definition of $\log$.

A first definition: One chooses a suitable simply connected domain $\Omega\subset {\mathbb C}\setminus\{0\}$ containing $1$ and defines $\log$ as $$\log z:=\int_1^z{d\zeta \over\zeta}\ .$$ Then it is obvious that ${d\over dz}\log z={1\over z}$.

Another way is the following: Choose a continuous branch ${\rm Arg}:\ \Omega\to{\mathbb R}\$ of the polar angle $\arg:\ {\mathbb R}^2\to{\mathbb R}/(2\pi{\mathbb Z})$ and put $$\log(x+iy)\ :=\ {1\over2}\log(x^2+y^2) + i\,{\rm Arg}(x,y)\ .$$ Then check by means of the CR equations that $\log$ is in fact an analytical function of the complex variable $z:=x+iy$.

In any case, restrictions on $\arg z$ only serve to make $\log$ well defined and have no effect on its derivative, which is always the function $z\mapsto {1\over z}$.

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The function $\log z$ is not continuous on any set containing a closed path around zero, so you need to restrict it to a subset of the plane that contains no such paths, for example, you "throw away" the ray from 0 to infinity directed $-\frac{\pi}{4}$ rad (which is the same as $\frac{7\pi}{4}$). So the reason for the restriction is making the function continuous and then, it's not hard to check that it'll be analytic too.
You can use the definition of the derivative once you know the function is analytic.

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I hope your answer satisfy the OP. A function is analytic if it is differentiable at the neighborhood of some point at which it is differentiable. So I do not think you have to check whether a function is analytic before applying the definition of derivative. –  Hassan Muhammad Mar 25 '12 at 6:30
@Hassan Muhammad: You are right, but the point is that if you apply the definition of the derivative on the function $\log z$ defined on $\mathbb{C}$ (clearly not continuously) you'll still be tempted to get $\frac{1}{z}$... –  Dennis Gulko Mar 25 '12 at 19:00