How do I derive:

$\ln(-1)=i\pi$ and

$\ln(-x)=\ln(x)+i\pi$ for $x>0$ and $x \in\mathbb R$

Thanks for any and all help!


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  • 7
    $\begingroup$ It is one of the values. $\endgroup$ – Will Jagy Nov 12 '14 at 21:46

See here to get justification: $$e^{i\pi} = -1$$ So $\ln(-1) := i\pi$ is reasonable. Note that $\ln(-1) = i(2k+1)\pi$ for some $k\in\mathbb Z$ is just as reasonable. It all comes down to the branch of $\ln$ wich is chosen.

For the second part enforce $\ln(ab) = \ln(a) + \ln(b)$ where $a = -1$ and $b=x$.


take the exponential to get $$e^{ln(-1)}=e^{\pi i}$$ $$-1=e^{\pi i}$$ $$e^{\pi i}+1=0$$ this is Euler's identity


What you said is not correct unless you select a branch. By definition, $log_{\theta}(-1):=ln|-1|+iarg_{\theta} (-1)= 0+iarg_{\theta}(-1)$ So you need $arg_{\theta}(-1)= \pi$. This is true for only one value $\theta$ (so that the branch goes from $\theta$ to $\theta + 2\pi^{-}$). So it seems like the branch with $ -\pi <arg\leq \pi$ should do it.

EDIT What you ultimately need in a branch log is to have the relation $$e^{logz}=z $$ , so that whatever argument $\theta$ you assign to z , you have $$ e^{i\theta}=cos(\theta)+isin(\theta)= i\pi$$. It then follows that $sin\theta= \pi$, so that $\theta =sin^{-1}(\pi)$

  • $\begingroup$ This is not entirely correct. We may chose the argument function such that $\arg(-1) = \pi$ as long as $(-\infty, 0)$ is inside our domain. $\endgroup$ – AlexR Nov 12 '14 at 22:27
  • $\begingroup$ @AlexR I don't understand what you believe is wrong in my answer. Of course, we cannot define an argument on a value that is not contained in the branch. Then if the argument is to be continuous, how else would you choose the branch so that $arg(-1)=\pi$ $\endgroup$ – Guest Nov 12 '14 at 22:35
  • $\begingroup$ I'm trying to point out that you can select an $\arg$ for any $B:=\mathbb C \setminus \langle e^{i\theta} \rangle_+$ such that $\ln:B\to\mathbb C$ satisfies $\ln(-1) = i\pi$, but you could also make it $\ln(-1) = 3\pi i$ on the same domain using a different (offset) $\arg$. $\arg_\theta$ isn't unique just by giving it's domain. $\endgroup$ – AlexR Nov 12 '14 at 22:39
  • $\begingroup$ But can you do the same without adding integer multiples of $i2\pi$ $\endgroup$ – Guest Nov 12 '14 at 22:41
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    $\begingroup$ You can, but that means 1. $\theta$ isn't unique (what you claimed), all you need is $\arg(B) \ni \pi$, wich is possible for any angular slice except $\theta = \pi + 2\pi k$ $\endgroup$ – AlexR Nov 12 '14 at 22:43

$e^{\pi i}=-1$ so $e^{\pi i}=i^2$ and $\ln(e^{\pi i})=\pi i$ and $\ln(i^2)=2\ln(i)$

Then $\pi i=2\ln(i)$. Note that $\pi=\frac{2\ln(i)}{i}$ and $e=i^{\frac1{\ln(i)}}$

$\ln(-x)=\ln(xi^2)$ so $\ln(-x)=\ln(x)+2\ln(i)$ or $\ln(-x)=\ln(x)+\pi i$


I'm always a little surprised people answer these with Euler's equation $e^{\pi i} = -1$ and expect it to be accepted without question that $e^{\pi i} = -1$.


We need to define what $e^{z}$ means if $z\in \mathbb C$. We want the basic properties that $e^x; x \in \mathbb R$ have to apply and carry over when we extend the definition of $e^z$ to allow complex numbers.

So first of all we want $e^{z + w} = e^z e^w$ which would mean we need for $z = x + yi$ where $x,y$ are real, sothat $e^{x + yi} = e^xe^{yi}$ which means we just need to define $e^{yi}$ somehow.

So we need to define $e^{yi} =E(y)$ for some $E:\mathbb R \to \mathbb C$ with $E(0) = 1$ (so that $e^{x} = e^{x + 0i} = e^xE(0)$.)

And second of all, we need the derivative of $e^z$ to equal $e^z$.

Some comments about derivatives of complex functions:

We can rewrite a complex function $f:x _ yi \mapsto u_{x,y} + v_{x,y} i$ as $f(z= x+yi) = u(x,y) + v(x,y)i$ where $u(x,y)$ and $v(x,y)$ are two real valued functions.

The derivative of $f(z) = \lim_{h\to 0}\frac {f(z+h) - f(z)}{h}$

If $h$ is real this is:

$f'(z) = \lim_{h\to 0} \frac {u(x+h,y) -u(x,y) +v(x+h,y)i - v(x,y)i}{h} = $

$\frac{du(x,y)}{dx} + \frac {dv(x,y)}{dx}i$.

But if $h=ik$ and is purely imaginary we have

$f'(z) = \lim_{k\to 0} \frac {u(x,y+k) -u(x,y) +v(x,y+k)i - v(x,y)i}{ik} = $

$f'(z) = \lim_{k\to 0} \frac {-u(x,y+k)i +u(x,y) i-v(x,y+k) - v(x,y)}{k} = $

$\frac {dv(x,y)}{dy} - \frac {du(x,y)}{dy}i$.

(Bear in mind that $\frac 1i = -i$ and $\frac ii = 1$)

For $f(z)$ to be differentiable (which suddenly we change the name of and call it analytical) we need the very restrictive condition:

$\frac {du(x,y)}{dx} = \frac {dv(x,y)}{dy}$ and $\frac {dv(x,y)}{dx} = -\frac {du(x,y)}{dy}$.

If we define $e^{z = x+iy} = e^xE(y)= u(x,y) + iv(x,y)$ where $u(x,y) = e^xRe(E(y)) + e^xIm(E(y))$. We can define $U(y)=Re(E(y))$ and $V(y)=Im(E(y))$ as real value functions where:

$u(x,y) = e^xU(y)$ and $v(x,y)=e^xV(y)$.

We need $\frac {du(x,y)}{dx} = \frac {de^x}{dx}U(y) + e^x\frac{dU(y)}{dx} = e^xU(y)$.

But also $\frac {du(x,y)}{dx}= \frac {dv(x,y)}{dy} = \frac {de^x}{dy}V(y) + e^x\frac {dV(y)}{dy} = e^xV'(y)$

While $\frac {dv(x,y)}{dx} = \frac {de^x}{dx}V(y) + e^x\frac{dV(y)}{dx} = e^xV(y)$.

But also $\frac {dv(x,y)}{dx} = -\frac {du(x,y)}{dy} = \frac {de^x}{dy}U(y) + e^x\frac {dU(y)}{dy} = e^xU'(y)$

So we have $U(y) = V'(y)$ and $V(y) = U'(y)$ and also (as $e^{x + 0i} =e^{x}(U(0) + V(0)i) = e^x$) we have $U(0) = 1$ and $V(0) = 0$.

$U(y) = \cos y$ and $V(y) = \sin y$ is one such pair of functions that satisfy those conditions.

I'll omit the details as to why thoes are the only pair of functions that satisfy those conditions.

So we define $e^{z = x+iy}=e^x(\cos y + \sin y i)$.


$e^{\pi i} = \cos \pi + \sin \pi i = -1$

Important caveat: $e^z$ is no longer $1-1$ so the "function" $\ln z$ is multivalued.

As $e^{z + 2k\pi i} = e^z $ , $\ln w = \ln w + 2k\pi i$. In other word $\ln w$ is a class of many complex numbers each differing from each other by a multiple of $2\pi i$


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