# What Is The Derivative Of Ln Z Derivative Of The Natural Logarithm

**What Is The Derivative Of Ln Z Derivative Of The Natural Logarithm**in

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I”ll use $log$ for the complex logarithm and $ln$ for the real-valued logarithm; you then have:$$log z = ln |z| + i arg z = ln r + ivarphi$$where I use $r = |z|$ and $varphi = arg z$ for simplicity. In this form, we have written$$log z = u(x,y)+iv(x,y)$$with $u(x,y) = ln r = ln sqrt{x^2+y^2} = frac{1}{2}ln(x^2+y^2)$ and $v(x,y) = varphi$.

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You can check the Cauchy-Riemann equations yourself and find (*)$$frac{partial u}{partial x} = frac{x}{x^2+y^2} = frac{partial varphi}{partial y} quad ; quad frac{partial u}{partial y} = frac{y}{x^2+y^2} = – frac{partial varphi}{partial x}$$Use this to find the derivative directly:$$frac{mbox{d} log z}{mbox{d}z} = frac{partial u}{partial x}+ifrac{partial v}{partial x} = frac{x-iy}{x^2+y^2} = frac{1}{z}$$

Remark: depending on how you *define* the complex logarithm, there will be different ways to find its derivative.

(*) The derivatives for $u$ are easy. For $v$, we have:

$$left{ egin{array}{ccc}x = rcosvarphi \y = rsinvarphiend{array}

ight.$$where both $r$ and $varphi$ are (implicit) functions of $x$ and $y$. Differentiating both equations w.r.t. $x$ gives:$$left{ egin{array}{ccc}1 = cosvarphifrac{partial r}{partial x}-rsinvarphifrac{partial varphi}{partial x} \0 = sinvarphifrac{partial r}{partial x}+rcosvarphifrac{partial varphi}{partial x}end{array}

ight.$$This is a linear system of two equations in the variables $frac{partial r}{partial x}$ and $frac{partial varphi}{partial x}$; solve for $frac{partial varphi}{partial x}$.

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In the same way, take the derivative w.r.t. $y$ and solve for $frac{partial varphi}{partial y}$. Can you take it from here?

Let $ U =mathbb C setminus (-infty,0>.$ Then $f(z)=ln |z| + iarg z$ is continuous on $U$ and we have $e^{f(z)} = z$ there. This shows $f(z)$ is injective on $U.$ Fix $zin U.$ Then for small nonzero $h$ we have

$$1=frac{e^{f(z+h)}-e^{f(z)}}{h} = frac{e^{f(z+h)}-e^{f(z)}}{f(z+h) – f(z)}frac{f(z+h) – f(z)}{h}.$$

The injectivity of $f$ shows that $f(z+h) – f(z)

e 0,$ so we”re OK dividing by it above. As $h o 0, f(z+h) o f(z)$ by the continuity of $f.$ So the first difference quotient on the right tends to the derivative of $e^w$ at $w=e^{f(z)},$ which is $e^{f(z)} = z

e 0.$ Knowing all of this, we can now write

$$ ag 1 frac{f(z+h) – f(z)}{e^{f(z+h)}-e^{f(z)}} = frac{f(z+h) – f(z)}{h}.$$

Since the left side of $(1) o 1/z,$ we get $f”(z) = 1/z$ (as expected).

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**Derivative**