There is a theorem that states that if $f$ is analytic in a domain $D$, and the closed disc {$ z:|z-\alpha|\leq r$} contained in $D$, and $C$ denotes the disc's boundary followed in the positive direction, then for every $z$ in the disc we can write: $$f(z)=\frac{1}{2\pi i}\int\frac{f(\zeta)}{\zeta-z}d\zeta$$

My question is: What is the intuitive explanation of this formula? (For example, but not necessary, geometrically.)

(Just to clarify - I know the proof of this theorem, I'm just trying to understand where does this exact formula come from.)


14 Answers 14


If you are looking for intuition then let us assume that we can expand $f(\zeta)$ into a power series around $z$: $f(\zeta) = \sum_{n \geq 0} c_n(\zeta - z)^n$. Note $c_0 = f(z)$. If you plug this into the integral and interchange the order of integration and summation then that integral on the right side of the formula becomes $\sum_{n \geq 0} \int c_n(\zeta - z)^{n-1}d\zeta$. Let us also assume that an integral along a contour doesn't change if we deform the contour continuously through a region where the function is "nice". So let us take as our path of integration a circle going once around the point $z$ (counterclockwise). Then you are basically reduced to showing that $\int (\zeta - z)^{m}d\zeta$ is 0 for $m \geq 0$ and is $2\pi i$ for $m = -1$. These can be done by direct calculations using polar coordinates with $\zeta = z + e^{it}$. Now divide by $2\pi i$ and you have the formula. Of course this is a hand-wavy argument in places, but the question was not asking for a rigorous proof. Personally, this is how I first came to terms with understanding how Cauchy's integral formula could be guessed.


The trick is to remember that differentiable is really a way of saying 'locally linear', that is, close to $z$ we have $$f( \zeta )= f(z) + (\zeta -z)f'(z) +g( \zeta )$$

Where $g(\zeta)$ is an error term that vanishes as we get close to $z$ (in fact it is $o(\zeta -z)$).

Now, dividing through by $(\zeta -z)$ as in the integral formula gives expression under the integral sign as: $$\frac{f( \zeta )}{(\zeta -z)}= \frac{f(z)}{(\zeta -z)} + f'(z) +\frac{g( \zeta )}{(\zeta -z)}$$

We now integrate this termwise:

The last term still vanishes as $\zeta \to z$ (by the definition of little o), which allows us to ignore it by way of a homotopy argument. You see, since we can shrink the loop $\gamma$ around $z$ arbitrarily small without changing the value of the integral (an old theorem of Cauchy's says that the integral along a closed curve is homotopy invariant $(*)$) and, as the loop gets smaller, we see that the value of the last term on $\gamma$ gets closer to vanishing- and so, morally, its integral over $\gamma$ may be bounded above in modulus by successively smaller $\epsilon$. But, because of homotopy invariance, these $\epsilon$ bound the integral over all such loops- and $\epsilon$ is arbitrary- so the integral of the last term vanishes.

The second term is easy- this integrates to $[\zeta f'(z)]_{\gamma (0)}^{\gamma (1)}$, with $\gamma (0)=\gamma (1)$. Plug it in to see this vanishes.

The first term is the interesting one and others have covered it well in their posts, but I'll finish the argument anyways. Having eliminated the other terms (and observing that $f(z)$ is a constant) we have a pair of statements equivalent to the original formula:

$$\frac{1}{2\pi i}\int\frac{f(z)}{\zeta-z}d\zeta =f(z) \iff \int\frac{1}{\zeta-z}d\zeta =2\pi i$$

This last one we can use substitution $u=\zeta-z$ and homotopy to make about the integral over a circle $\gamma (t)= e^{2\pi it}$ around the origin giving:

$$\int_{\gamma} \frac{1}{u} du (**)= \int_0^1 \frac{\gamma '(t)}{\gamma (t)}dt= = \int_0^1 \frac{2 \pi i e^{2\pi it}}{e^{2\pi it}}dt$$

The last one is a total gift after cancellation and we all go home smiling. We could of course have morally calculated $(**)$ without resorting to technicalities by observing that the primitive of $\frac{1}{u}= log(u)$ and looking at the integrand $[log(u)]_{e^0}^{e^{(2\pi -\epsilon) i}}$ and letting $\epsilon \to 0$. But the 'right' intuition is certainly that of Augusti's answer.

As for $(*)$, the intuition is another story altogether, but I may add it later if it looks worth it...


As Chandru1 pointed out, a beautiful particular case of Cauchy's integral formula gives us a very particular winding number. For the sake of simplicity, let us look only at closed paths on the boundary of the unit disc, $\gamma : [0,1] \longrightarrow S^1$. Then, the winding number of $\gamma$ is the number of times $\gamma$ goes round the circumference. That is, if we write the complex number $\gamma (t)$ in its exponential form

$$ \gamma (t) = e^{i\theta (t)} \qquad \qquad \qquad [1] $$

then you can prove that there is a continuous choice for the argument $\theta (t)$ of $\gamma (t)$; that is, a continuous function $\theta : [0, 1] \longrightarrow \mathbb{R} $ such that [1] holds for all $t$. In Algebraic Topology, $\theta$ is called a "lifting" of $\gamma$. Moreover, any two such continuous "liftings" of $\gamma$ differ necessarily by a constant integer multiple of $2\pi$: if $\widetilde{\theta}(t)$ is another (continuous) lifting of $ \gamma$, then there is a constant integer $k$, not depending on $t$, such that $\widetilde{\theta}(t) = \theta (t) + 2k\pi$ for all $t$. (This is called the "lifting lemma" in Algebraic Topology. For a proof in Complex Analysis, see theorem 7.1, in Stewart and Tall.) So the number of times $\gamma $ goes round the circumference (its winding number) is well-defined as

$$ w (\gamma , 0) = \frac{\theta (1) - \theta (0)}{2\pi} \ . $$

Then you can easily prove (see op.cit, section 7.5), that

$$ w (\gamma , 0 ) = \frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}d\zeta \ . $$

So, Cauchy's integral formula for the constant function $f \equiv 1$, $z = 0$ (and $\gamma (t) = e^{i2\pi t}$, hence $\theta (t) =2 \pi t$) tells us that

$$ \frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}d\zeta = 1 \ . $$

That is, if you go round the circumference just one time, well, you are going round the circumference indeed exactly once. Isn't math amazing? :-)

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    $\begingroup$ (OT) Agusti, apologies for the edit war; Chandru, it is more polite to just point out what you think is a mistake and let the writer fix it him/herself. $\endgroup$ Sep 5, 2010 at 15:33
  • $\begingroup$ @J.M. I was just wondering myself what was going on with that twinkling "aren't / isn't". Anyway, guys, I assume you are the native English speakers, not me, and I appreciate both of your efforts to improve my poor English, but if "maths" is plural, then it should be "aren't", not "isn't", isn't it? :-) $\endgroup$ Sep 5, 2010 at 15:39
  • $\begingroup$ And that's my reason for the rollback. :) $\endgroup$ Sep 5, 2010 at 15:59
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    $\begingroup$ My Webster says "mathematics, n pl but usu sing in constr"... $\endgroup$ Sep 5, 2010 at 17:20
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    $\begingroup$ Dear Agusti, As you probably know, "maths" is used in Britain, Australia, and most other non-US English speaking countries (except Canada, maybe?), and "math" is used in the US. Even though "maths" ends in "s", we still use singular constructions with it (and the same with the unabbreviated form "mathematics"). (I say this with the authority of having been born and grown up in Australia.) I can certainly see that it looks confusing to a non-native speaker! $\endgroup$
    – Matt E
    Sep 8, 2010 at 5:08

Cauchy's Formula has a remarkable interpretation in terms of hyperbolic geometry.

To understand it, you need to know very little about hyperbolic geometry.

In fact, you only need to know that if you define the "plane" as the open unit disk $D$ in $\mathbb C$, and the "lines" as those arcs of circle inside $D$ which are orthogonal to the unit circle $\partial D$, you get a model for the hyperbolic plane.

[There is a first miracle: the orientation preserving isometries of $D$ into itself are precisely the holomorphic automorphisms of $D$.]

If $f$ is holomorphic in a neighborhood of the closure of $D$, and $z$ is in $D$, then Cauchy's Formula says that

$f(z)$ is the average of $f$ on $\partial D$ "as you see it from $z$".

By this I mean that $f(z)$ is the integral of $f$ on $\partial D$ with respect to the measure which assigns to an arc in $\partial D$ the number $\theta/2\pi$ where $\theta\in[0,2\pi]$ is the angle between the (hyperbolic) half-lines going from $z$ to the end points of the arc. (That's the measure you think the arc has, as a part of your horizon, if you look at it from $z$.)

EDIT 1 OF NOV 1, 2010.

This is to clarify the relationship between the Cauchy's and Poisson's Formulas.

For $|z|<1$ and $\theta$ real define the Cauchy kernel by $$C(\theta,z):=\frac{1}{2\pi}\ \frac{e^{i\theta}}{e^{i\theta}-z}\quad,$$ and define the Cauchy transform of a continuous function $g$ on the unit circle $\partial D$ by $$(Cg)(z):=\int_0^{2\pi}g(e^{i\theta})\ C(\theta,z)\ d\theta$$ for $|z|<1$. In particular Cauchy's Formula says that if $g$ is the boundary value of a holomorphic function $f$ on $D$, then $Cg=f$.

For $|z|<1$ and $\theta$ real define the Poisson kernel by $$P(\theta,z):=C(\theta,z)+\overline{C(\theta,z)}-\frac{1}{2\pi}\quad,$$ and define the Poisson transform of a continuous function $g$ on the unit circle $\partial D$ by $$(Pg)(z):=\int_0^{2\pi}g(e^{i\theta})\ P(\theta,z)\ d\theta$$ for $|z|<1$. In particular Poisson's Formula says that if $g$ is the boundary value of a harmonic function $f$ on $D$, then $Pg=f$.

EDIT 2 OF NOV 1, 2010. I stole this from Bill Thurston: go to page 180 of (or search for "visual" in) http://www.math.unl.edu/~mbrittenham2/classwk/990s08/public/thurston.notes.pdf/8a.pdf .

EDIT OF NOV 2, 2010.

Stricto sensu, there is no geometric interpretation of the Cauchy kernel, because it is not invariant.

Indeed, if $G$ denotes the group of biholomorphic transforms of the open unit disk $D$, then the Cauchy kernel $C(z,\theta)d\theta$, viewed as a 1-form on $D\times\partial D$, is not $G$-invariant.

However, its restriction to $\{0\}\times\partial D$ is invariant under the stabilizer of $0\in D$ in $G$ (which is the circle group).

As a result, this restriction extends in a unique way to a $G$-invariant 1-form on $D\times\partial D$, and this 1-form is the Poisson kernel.

More precisely, the (complex) vector space of $G$-invariant $(0,1)$-forms on $D\times\partial D$ is one dimensional, generated by the Poisson kernel. [This is because the action of $G$ on $D\times\partial D$ is simply transitive, and the $(0,1)$-forms on $D\times\partial D$ are the sections of a homogeneous line bundle over $D\times\partial D$.]

  • $\begingroup$ While the hyperbolic geometry interpretation is true for a harmonic function $f : D \to \mathbb{R}$, I contest that it is a good interpretation of Cauchy's Formula. Sure, real and imaginary part of an analytic function are harmonic, but the trouble is that the $1/(z-a)$ factor mixes the real and the imaginary part. In other words, it assign a complex measure to the arc $C$, not just a real arc length. How do you untangle this? $\endgroup$ Oct 31, 2010 at 21:54
  • $\begingroup$ @Greg Graviton - Thanks for your comment. Unfortunately I don't understand it. Sorry. [By the way, I'm not claiming that my point is original.] $\endgroup$ Nov 1, 2010 at 5:40
  • $\begingroup$ @Pierre-Yves. :D Let me try again. (Ah, I messed up notation.) You are saying that $f(z) = \int_{\partial D} f(\zeta) d\theta$ where $d\theta$ is the measure that associates to each arc $C$ the angle between hyperbolic half-lines emanating from $z$. But Cauchy's Formula asserts that $d\theta = \frac{\zeta}{\zeta - z}d\phi$ where $d\phi$ is the ordinary (Haar) measure on the circle $\partial D$ (and $\zeta=re^{i\phi}$). The trouble is that the latter measure is complex valued, while the former one is real valued; I don't see how your $d\theta$ is a good substitute for Cauchy's $d\theta$. $\endgroup$ Nov 1, 2010 at 8:08
  • $\begingroup$ @Greg Graviton - I think I'm beginning to get your point. I agree with you that the "visual measure at z" (i.e. the unique probability measure on the circle which is invariant under the stabilizer of z) is NOT given by the Cauchy kernel. It's given by the Poisson kernel. But the Cauchy transform and the Poisson transform coincide on boundary values of holomorphic functions. $\endgroup$ Nov 1, 2010 at 8:49
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    $\begingroup$ @Pierre-Yves. Yep. Now, my point is that it is far from obvious or intuitive that the Poisson transform and the Cauchy transform coincide for analytic functions! (To me at least. If you have a nice argument, please tell!) (Considering the real and imaginary part of $f$ separately doesn't work.) $\endgroup$ Nov 1, 2010 at 11:20

In short, the intuitive reason the Cauchy integral formula is true is that (a) it is true in the limit for small circle paths $C_r$ about $z$, when it reduces to averaging, and (b) the integral does not change as the path deforms nicely from $C_r$ to $C$:

  1. Integrals of complex differentiable functions $\zeta\mapsto F(\zeta)$ over closed paths don't change when the paths are deformed nicely in the domain of differentiability. For a super proof of this that does not use continuous differentiability, see H. Hanche-Olsen, ``On Goursat's proof of the Cauchy integral theorem,'' Amer. Math. Monthly 115 (2008) 648-652.

  2. Integration on circle paths $C_r: t\mapsto \zeta=z+re^{it}$ ($t\in[0,2\pi]$) corresponds to averaging: $$ \frac{d\zeta}{\zeta-z} = \frac{i r e^{it} dt}{re^{it}} =i\,dt, \quad\mbox{hence}\ \frac{1}{2\pi i}\int_{C_r} \frac{f(\zeta)}{\zeta-z}d\zeta = \frac1{2\pi} \int_0^{2\pi}f(z+re^{it})\,dt. $$

  3. Given $z\in D$, the function $$\zeta\mapsto F(\zeta)=\frac{f(\zeta)}{\zeta-z}$$ is differentiable in $D\setminus\{z\}$, so for any path, like $C$, that can be deformed nicely in $D\setminus\{z\}$ to a circle path $C_r$ as above, we can further deform by taking $r\to0$ and conclude that for all small positive $r$, $$ f(z) =\frac{1}{2\pi i} \int_{C_r} \frac{f(\zeta)}{\zeta-z}d\zeta =\frac{1}{2\pi i} \int_C \frac{f(\zeta)}{\zeta-z}d\zeta. $$

  • $\begingroup$ Very nice answer. $\endgroup$ Mar 25, 2011 at 13:59

An exercise that was a big help to me was to compute the integral of 1/z on the closed path consisting of the square from (1,-1) to (1,1) to (-1,1) to (-1,-1) to (1,-1). It illustrates how the sides with varying real differ from the sides with varying imaginary values.

  • $\begingroup$ great idea ! thx ! $\endgroup$ Nov 24, 2018 at 20:33

See Needham's Visual Complex Analysis for lots of geometric intuition and much more.

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    $\begingroup$ This is a great book for intuition! $\endgroup$
    – Henry
    Dec 14, 2022 at 21:40

I don't have enough reputation to add a comment, but this is just a minor add-on to KCd's answer. He points out that the whole thing reduces to computing integrals of the form $\int z^m dz$, where m is an integer. Certainly one can compute these integrals by polar coordinates, but the following way to think about this computation is in my opinion closer to geometric intuition.

Namely, when $m \neq -1$ the function $z^m$ has a single-valued antiderivative $z^{m+1}/(m+1)$ on $\mathbb C \setminus \{0\}$, and so the integral of $z^m$ over any closed loop in $\mathbb C \setminus \{0\}$ vanishes, even though the punctured plane is not simply connected. The only m for which one fails to get a single-valued antiderivative is $m=-1$, in which case one gets the multi-valued function $\log z$. The value of $\log z$ for two ''adjacent'' branches always differs by exactly $2\pi i$, so the integral of $1/z$ over a closed loop in $\mathbb C \setminus \{0\}$ just gives you $2\pi i$ times the winding number of the loop around the origin, i.e. the total number of times it has ''jumped'' to another branch in the counterclockwise direction. In the assumptions of your theorem the boundary C will traverse the singularity once counterclockwise, which gives the result.

  • $\begingroup$ This is very insightful, thank you. Makes sense that this one rational function (1/x) whose integral is not a rational function but a multivalued logarithm, makes this special residual. I was wondering if there's a physical interpretation based on alternating current or something else where the area in the complex plane has some meaning? $\endgroup$
    – keithphw
    Aug 8, 2023 at 0:51

Expanding on my comment, this result can be translated into:

"A surface in $\mathbb{R}^3$ which satisfies the Maximum-Modulus principle is uniquely determined by specifying it's boundary"

To see this, write the holomorphic function $f(z)$ in terms of its real and imaginary parts:

$$ f(z) = f(x,y) = g(x,y) + ih(x,y)$$

Then since $f$ is holomorphic, the functions $g(x,y)$ and $h(x,y)$ are both real valued harmonic functions which satisfy the maximum modulus principle. By interpreting the value of $g$ or $h$ as the height of a surface in $\mathbb{R}^3$, we can see that according to Cauchy's theorem such surfaces are uniquely determined by specifying their boundary.

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    $\begingroup$ Can you elaborate please I would like to understand how that can be! $\endgroup$
    – anon
    Sep 14, 2010 at 19:54

Perhaps you can see the Cauchy's integral formula as the "Winding number" of a curve $C$ around any complex number $a$.

For more information, please see : en.wikipedia.org/wiki/Winding_number


Even though I am not totally satisfied by my argument, I would say it is about average values on the curve of integration that is dedicated to analytic functions. Lets look at a circle $C$ around $z$: If we choose points $\zeta_1,\ldots,\zeta_n$ on $C$ such that they span a regular $n$-sided polygon the average of their sum is $z$, that is $$\sum_{k=1}^n \zeta_k\cdot\frac{1}{n}= z.$$

This is far from Cauchy integral formula, however, since $$1 = \frac{1}{2\pi i}\int_C \frac{1}{\zeta-z}d\zeta$$ we can write the Cauchy formula in the form $$\frac{1}{2\pi i}\int_C\frac{f(\zeta)-f(z)}{\zeta-z}d\zeta=0.$$ Next, if we think of $f$ as a power series around $z$, we can interpret the formula in text as saying that:

"If we remove the constant term $f(z)$ from $f(\zeta)$ and divide out the factor $\zeta-z$, the remaining will have average 0".


I am learning Cauchy's Integral theorem for the first time too and have a very simple and perhaps incorrect physical interpretation which allows me to memorize this rule and use it... for now...I'll need a better understanding to be able to sleep at night once I pass the course :)

In fluid mechanics, we often care about the flow (velocity field) in a plane. This flow be may idealized as a uniform flow, where all velocity vectors obey the Cauchy Riemann relations and the function that describes this flow is analytic. T

Lets imagine that in a plane, there is a curl free flow, from a sourse that exists in a third dimension which we can't see. The source has a function, but it looks like a discontinuity on the plane. Lets say that water pours onto the plan, spreading out evenly in all directions, all divergence no curl.

This flow is analytic and the integral around it (stokes theorem) is going to be zero everywhere but at the singularity. How much water is diverging...? The amount coming from the source, or the function f(z_0).

If this is total BS please correct...


We want to show: $$ 2\pi i f(a) =\oint_{S^1} \frac{f(z) dz}{z-a} $$ where $S^1$ is the circle paramaterized by $\gamma(t)=a+re^{it}$. Putting $z=\gamma(t)=a+re^{it}$, we get $$ \frac {dz} {z-a} = \frac{ire^{it}}{re^{it}} dt = idt $$ Hence $$ \oint_{S^1} \frac{f(z)dz}{z}=\int_0^{2\pi}f(re^{it})idt $$ As $r\to 0$, the above becomes $f(a)$

  • $\begingroup$ The OP asked for an intuitive explanation. $\endgroup$ Sep 29, 2015 at 19:30

The reason we can change/deform contours is due to a generalized version of Stoke's Theorem. Which has an extremely intuitive explanation here on Wikipedia, see underlying principles. Here is another here. The hand wavy reason it works is because the curl of a complex function is $0$.

Circles are really easy to parametrize so that's why we (generally) integrate about circles rather than..say triangles. This is due to Euler's Identity.

So when we integrate about a curve, $C(\theta)$, we multiply by a differential $dC(\theta)$ such that

$$(1) \quad dC(\theta)=C(\theta+d\theta)-C(\theta)$$

Multiplying and dividing the right hand side of $(1)$ by $d\theta$ yeilds,

$$(2) \quad dC(\theta)= \cfrac{dC}{d\theta} \cdot dt$$


$$(3) \quad \int_C f(C(\theta)) \ dC(\theta)=\int f(C(\theta)) \cdot \cfrac{dC}{d\theta} \ d\theta$$

The meaning of $(3)$ is as follows. The Line Integral of a function $f(x)$ about a curve $C(\theta)$ gives the average value of $f \cdot \cfrac{dC}{|dC|}$ on that curve multiplied by the length of the curve.

If you accept this claim, we can simply substitute the functions in question, however that isn't really the issue is it?

First, let's address a major issue; why does $(3)$ average over $f \cdot \cfrac{dC}{|dC|}$?

It'll take some imagination but a reasonable explanation can be given. Imagine $f(t)$ gives the speed of a walking person at a particular time. If it's positive, (s)he's walking to the right. If it's negative, (s)he's walking to the left. Now imagine that we were viewing this on a TV rather than in real life. If we speed up time, fast forwarded, the person would appear to be walking faster. If we rewinded te person would switch directions. However, if we have a real time $p$ then we can know the time on the TV by making $t$ a function of the real time $p$. Now, if $dt$ is negative, we know that the time on the TV is going backward. However, if the person's speed is still positive, we know that the true speed, relative to the observer at time $p$, is actually negative. So if we want the average speed of the person, it's not enough to watch the TV, you also need to know how time is progressing. That's why $(3)$ can be negative. However, if time progresses normally relative to the observer, the ratio becomes unity.

So, in short $(3)$ averages, but does so with respect to direction. Sadly, explaining imaginary time, would take a lot of real time, so we'll have to settle for the above sentence rather than the time analogy.

However, recall that multiplication by a complex number $v$ rotates the multiplicand $w$ by $arg(v)$ degrees/radians and scales by $|v|$ in the complex plane. For a more in depth explanation and review see here.

So, $\cfrac{dC(\theta)}{|dC(\theta)|}$ gives the normed differential, the direction of the differential, at an angle $\theta$.

Imagine lines running from the origin to a point on $C$. Then imagine another line going from that point then tangent counter-clockwise along the curve. Move this new line and place it's tail at the origin. So, you should see a line representing $C(t)$, then you should also see another line representing the direction of $dC$. Notice that for the complex circle, $90^o$ seperates these lines. Since, multiplication can rotate, the direction of $dC$ is $C$ rotated by $90^o$. Better yet,

$(4) \quad dC=i \cdot C$

$\Rightarrow arg(dC)=arg(C)+\pi/2$

Where we've switched to radians. It's important to realize that it's the difference between the directions, angles, that's key. In symbols,

$(5) \quad \Delta \theta=arg(dC)-arg(C)=\pi/2$

Which is constant. We know that,

$(6) \quad arg(v \cdot w)=arg(w)+arg(v)$

By extension,

$(7) \quad arg \left(\cfrac{w}{v} \right)=arg(w)-arg(v)$

Using the principle of correspondence yeilds,

$(8) \quad arg \left(\cfrac{dC}{C} \right)=arg(dC)-arg(C)$

Thus, if we want $(3)$ to be constant, if we want the angle difference to be constant, we should have $f=\cfrac{1}{C(\theta)}$.

What about $f=\cfrac{1}{C(\theta)^2}$? It rotates faster than $dC$, so the difference between angles won't be constant.

What about $f=\cfrac{1}{C(\theta)^0}$? This time it rotates to slowly for the difference to be constant.

Finally, putting everything together,

$$(9) \quad \cfrac{1}{2 \cdot \pi} \cdot \int_{C_r} \cfrac{dC}{C-z_0}=i$$

This says that the average angle between $dC$ and $\cfrac{1}{C-z_0}$ in the complex plane is $\pi/2$ radians or $90^o$ degrees. Which in this case is best represented as $i$.

With this inuition it's now easy to see how Cauchy's Integral Formula works in general.

(Reproduced with permission from here)


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