Inmathematics, thegamma function (represented by Γ, capitalGreek lettergamma) is the most common extension of thefactorial function tocomplex numbers. Derived byDaniel Bernoulli, the gamma function${\displaystyle \mathrm{\Gamma }(z)}$ is defined for all complex numbers${\displaystyle z}$ except non-positive integers, and for everypositive integer${\displaystyle z=n}$,$${\displaystyle \mathrm{\Gamma }(n)=(n-1)!.}$$The gamma function can be defined via a convergentimproper integral for complex numbers with positive real part:

Gamma
The gamma function along part of the real axis
General information
General definition	${\displaystyle \mathrm{\Gamma }(z)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}dt}$
Fields of application	Calculus, mathematical analysis, statistics, physics

$${\displaystyle \mathrm{\Gamma }(z)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}\text{ d}t,\mathrm{\Re }(z)>0.}$$The gamma function then is defined in the complex plane as theanalytic continuation of this integral function: it is ameromorphic function which isholomorphic except at zero and the negative integers, where it has simplepoles.

The gamma function has no zeros, so thereciprocal gamma function⁠1/Γ(z)⁠ is anentire function. In fact, the gamma function corresponds to theMellin transform of the negativeexponential function:

$${\displaystyle \mathrm{\Gamma }(z)=\mathcal{M}\{e^{-x}\}(z).}$$

Other extensions of the factorial function do exist, but the gamma function is the most popular and useful. It appears as a factor in various probability-distribution functions and other formulas in the fields ofprobability,statistics,analytic number theory, andcombinatorics.

The gamma function can be seen as a solution to theinterpolation problem of finding asmooth curve${\displaystyle y=f(x)}$  that connects the points of the factorial sequence:${\displaystyle (x,y)=(n,n!)}$  for all positive integer values of${\displaystyle n}$ . The simple formula for the factorial,x! = 1 × 2 × ⋯ ×x is only valid whenx is a positive integer, and noelementary function has this property, but a good solution is the gamma function${\displaystyle f(x)=\mathrm{\Gamma }(x+1)}$ .^[1]

The gamma function is not only smooth butanalytic (except at the non-positive integers), and it can be defined in several explicit ways. However, it is not the only analytic function that extends the factorial, as one may add any analytic function that is zero on the positive integers, such as${\displaystyle k\sin (m\pi x)}$  for an integer${\displaystyle m}$ .^[1] Such a function is known as apseudogamma function, the most famous being theHadamard function.^[2]

A more restrictive requirement is thefunctional equation which interpolates the shifted factorial${\displaystyle f(n)=(n-1)!}$  :^[3][4]$${\displaystyle f(x+1)=xf(x)\text{ for all }x>0,f(1)=1.}$$ 

But this still does not give a unique solution, since it allows for multiplication by any periodic function${\displaystyle g(x)}$  with${\displaystyle g(x)=g(x+1)}$  and${\displaystyle g(0)=1}$ , such as${\displaystyle g(x)=e^{k\sin (m\pi x)}}$ .

One way to resolve the ambiguity is theBohr–Mollerup theorem, which shows that${\displaystyle f(x)=\mathrm{\Gamma }(x)}$  is the unique interpolating function for the factorial, defined over the positive reals, which islogarithmically convex,^[5] meaning that${\displaystyle y=\log f(x)}$  isconvex.^[6]

The notation${\displaystyle \mathrm{\Gamma }(z)}$  is due toLegendre.^[1] If the real part of the complex number z is strictly positive (${\displaystyle \mathrm{\Re }(z)>0}$ ), then theintegral$${\displaystyle \mathrm{\Gamma }(z)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}dt}$$ converges absolutely, and is known as theEuler integral of the second kind. (Euler's integral of the first kind is thebeta function.^[1]) Usingintegration by parts, one sees that:

 

Recognizing that${\displaystyle -t^z{e}^{-t}\to 0}$  as${\displaystyle t\to \mathrm{\infty },}$  

Then${\displaystyle \mathrm{\Gamma }(1)}$  can be calculated as: 

Thus we can show that${\displaystyle \mathrm{\Gamma }(n)=(n-1)!}$  for any positive integern byinduction. Specifically, the base case is that${\displaystyle \mathrm{\Gamma }(1)=1=0!}$ , and the induction step is that${\displaystyle \mathrm{\Gamma }(n+1)=n\mathrm{\Gamma }(n)=n(n-1)!=n!.}$ 

The identity$\mathrm{\Gamma }(z)=\frac{\mathrm{\Gamma }(z+1)}{z}$  can be used (or, yielding the same result,analytic continuation can be used) to uniquely extend the integral formulation for${\displaystyle \mathrm{\Gamma }(z)}$  to ameromorphic function defined for all complex numbersz, except integers less than or equal to zero.^[1] It is this extended version that is commonly referred to as the gamma function.^[1]

There are many equivalent definitions.

For a fixed integer${\displaystyle m}$ , as the integer${\displaystyle n}$  increases, we have that^[7]$${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{n!{\left(n+1\right)}^m}{(n+m)!}=1.}$$ 

If${\displaystyle m}$  is not an integer then this equation is meaningless since, in this section, the factorial of a non-integer has not been defined yet. However, let us assume that this equation continues to hold when${\displaystyle m}$  is replaced by an arbitrary complex number${\displaystyle z}$ , in order to define the Gamma function for non-integers:

$${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{n!{\left(n+1\right)}^z}{(n+z)!}=1.}$$ Multiplying both sides by${\displaystyle (z-1)!}$  gives Thisinfinite product, which is due to Euler,^[8] converges for all complex numbers${\displaystyle z}$  except the non-positive integers, which fail because of a division by zero. In fact, the above assumption produces a unique definition of${\displaystyle \mathrm{\Gamma }(z)}$  as⁠${\displaystyle (z-1)!}$ ⁠.

Intuitively, this formula indicates that${\displaystyle \mathrm{\Gamma }(z)}$  is approximately the result of computing${\displaystyle \mathrm{\Gamma }(n+1)=n!}$  for some large integer${\displaystyle n}$ , multiplying by${\displaystyle (n+1{)}^z}$  to approximate${\displaystyle \mathrm{\Gamma }(n+z+1)}$ , and then using the relationship${\displaystyle \mathrm{\Gamma }(x+1)=x\mathrm{\Gamma }(x)}$  backwards${\displaystyle n+1}$  times to get an approximation for${\displaystyle \mathrm{\Gamma }(z)}$ ; and furthermore that this approximation becomes exact as${\displaystyle n}$  increases to infinity.

The infinite product for thereciprocal$${\displaystyle \frac{1}{\mathrm{\Gamma }(z)}=z\prod _{n=1}^{\mathrm{\infty }}\left[\left(1+\frac{z}{n}\right)/{\left(1+\frac{1}{n}\right)}^z\right]}$$ is anentire function, converging for every complex numberz.

The definition for the gamma function due toWeierstrass is also valid for all complex numbers ${\displaystyle z}$  except non-positive integers:$${\displaystyle \mathrm{\Gamma }(z)=\frac{e^{-\gamma z}}{z}\prod _{n=1}^{\mathrm{\infty }}{\left(1+\frac{z}{n}\right)}^{-1}{e}^{z/n},}$$ where${\displaystyle \gamma \approx 0.577216}$  is theEuler–Mascheroni constant.^[1] This is theHadamard product of${\displaystyle 1/\mathrm{\Gamma }(z)}$  in a rewritten form.

Besides the fundamental property discussed above:$${\displaystyle \mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }(z)}$$ other important functional equations for the gamma function areEuler's reflection formula$${\displaystyle \mathrm{\Gamma }(1-z)\mathrm{\Gamma }(z)=\frac{\pi }{\sin \pi z},z\notin \mathbb{Z}}$$ which implies$${\displaystyle \mathrm{\Gamma }(z-n)=(-1{)}^{n-1}\frac{\mathrm{\Gamma }(-z)\mathrm{\Gamma }(1+z)}{\mathrm{\Gamma }(n+1-z)},n\in \mathbb{Z}}$$ and theLegendre duplication formula$${\displaystyle \mathrm{\Gamma }(z)\mathrm{\Gamma }\left(z+\frac{1}{2}\right)=2^{1-2z}\sqrt{\pi }\mathrm{\Gamma }(2z).}$$ 

The duplication formula is a special case of themultiplication theorem (see ^[9] Eq. 5.5.6):$${\displaystyle \prod _{k=0}^{m-1}\mathrm{\Gamma }\left(z+\frac{k}{m}\right)=(2\pi {)}^{\frac{m-1}{2}}{m}^{\frac{1}{2}-mz}\mathrm{\Gamma }(mz).}$$ 

A simple but useful property, which can be seen from the limit definition, is:$${\displaystyle \overline{\mathrm{\Gamma }(z)}=\mathrm{\Gamma }(\overline{z})\Rightarrow \mathrm{\Gamma }(z)\mathrm{\Gamma }(\overline{z})\in \mathbb{R}.}$$ 

In particular, withz =a +bi, this product is$${\displaystyle |\mathrm{\Gamma }(a+bi){|}^2=|\mathrm{\Gamma }(a){|}^2\prod _{k=0}^{\mathrm{\infty }}\frac{1}{1+\frac{b^2}{(a+k{)}^2}}}$$ 

If the real part is an integer or a half-integer, this can be finitely expressed inclosed form: 

Perhaps the best-known value of the gamma function at a non-integer argument is$${\displaystyle \mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi },}$$ which can be found by setting$z=\frac{1}{2}$  in the reflection formula, by using the relation to thebeta function given below with$z_1=z_2=\frac{1}{2}$ , or simply by making the substitution${\displaystyle t=u^2}$  in the integral definition of the gamma function, resulting in aGaussian integral. In general, for non-negative integer values of${\displaystyle n}$  we have: where thedouble factorial${\displaystyle (2n-1)!!=(2n-1)(2n-3)\cdots (3)(1)}$ . SeeParticular values of the gamma function for calculated values.

It might be tempting to generalize the result that$\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }$  by looking for a formula for other individual values${\displaystyle \mathrm{\Gamma }(r)}$  where${\displaystyle r}$  is rational, especially because according toGauss's digamma theorem, it is possible to do so for the closely relateddigamma function at every rational value. However, these numbers${\displaystyle \mathrm{\Gamma }(r)}$  are not known to be expressible by themselves in terms of elementary functions. It has been proved that${\displaystyle \mathrm{\Gamma }(n+r)}$  is atranscendental number andalgebraically independent of${\displaystyle \pi }$  for any integer${\displaystyle n}$  and each of the fractions$r=\frac{1}{6},\frac{1}{4},\frac{1}{3},\frac{2}{3},\frac{3}{4},\frac{5}{6}$ .^[10] In general, when computing values of the gamma function, we must settle for numerical approximations.

The derivatives of the gamma function are described in terms of thepolygamma function, ψ⁽⁰⁾(z):$${\displaystyle {\mathrm{\Gamma }}^{\prime }(z)=\mathrm{\Gamma }(z){\psi }^{(0)}(z).}$$ For a positive integer m the derivative of the gamma function can be calculated as follows:

$${\displaystyle {\mathrm{\Gamma }}^{\prime }(m+1)=m!\left(-\gamma +\sum _{k=1}^m\frac{1}{k}\right)=m!\left(-\gamma +H(m)\right),}$$ where H(m) is the mthharmonic number andγ is theEuler–Mascheroni constant.

For${\displaystyle \mathrm{\Re }(z)>0}$  the${\displaystyle n}$ th derivative of the gamma function is:$${\displaystyle \frac{d^n}{d{z}^n}\mathrm{\Gamma }(z)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}(\log t{)}^{n}dt.}$$ (This can be derived by differentiating the integral form of the gamma function with respect to${\displaystyle z}$ , and using the technique ofdifferentiation under the integral sign.)

Using the identity$${\displaystyle {\mathrm{\Gamma }}^{(n)}(1)=(-1{)}^n{B}_n(\gamma ,1!\zeta (2),\dots ,(n-1)!\zeta (n))}$$ where${\displaystyle \zeta (z)}$  is theRiemann zeta function, and${\displaystyle B_n}$  is the${\displaystyle n}$ -thBell polynomial, we have in particular theLaurent series expansion of the gamma function^[11]$${\displaystyle \mathrm{\Gamma }(z)=\frac{1}{z}-\gamma +\frac{1}{2}\left({\gamma }^2+\frac{{\pi }^2}{6}\right)z-\frac{1}{6}\left({\gamma }^3+\frac{\gamma {\pi }^2}{2}+2\zeta (3)\right)z^2+O(z^3).}$$ 

When restricted to the positive real numbers, the gamma function is a strictlylogarithmically convex function. This property may be stated in any of the following three equivalent ways:

• For any two positive real numbers${\displaystyle x_1}$  and${\displaystyle x_2}$ , and for any${\displaystyle t\in [0,1]}$ ,$${\displaystyle \mathrm{\Gamma }(t{x}_1+(1-t)x_2)\le \mathrm{\Gamma }(x_1{)}^t\mathrm{\Gamma }(x_2{)}^{1-t}.}$$ 
• For any two positive real numbers${\displaystyle x_1}$  and${\displaystyle x_2}$ , and${\displaystyle x_2}$  >${\displaystyle x_1}$ $${\displaystyle {\left(\frac{\mathrm{\Gamma }(x_2)}{\mathrm{\Gamma }(x_1)}\right)}^{\frac{1}{x_2-x_1}}>\exp \left(\frac{{\mathrm{\Gamma }}^{\prime }(x_1)}{\mathrm{\Gamma }(x_1)}\right).}$$ 
• For any positive real number${\displaystyle x}$ ,$${\displaystyle {\mathrm{\Gamma }}^{″}(x)\mathrm{\Gamma }(x)>{\mathrm{\Gamma }}^{\prime }(x{)}^2.}$$ 

The last of these statements is, essentially by definition, the same as the statement that${\displaystyle {\psi }^{(1)}(x)>0}$ , where${\displaystyle {\psi }^{(1)}}$  is thepolygamma function of order 1. To prove the logarithmic convexity of the gamma function, it therefore suffices to observe that${\displaystyle {\psi }^{(1)}}$  has a series representation which, for positive realx, consists of only positive terms.

Logarithmic convexity andJensen's inequality together imply, for any positive real numbers${\displaystyle x_1,\dots ,x_n}$  and${\displaystyle a_1,\dots ,a_n}$ ,$${\displaystyle \mathrm{\Gamma }\left(\frac{a_1{x}_1+\cdots +a_n{x}_n}{a_1+\cdots +a_n}\right)\le (\mathrm{\Gamma }(x_1{)}^{a_1}\cdots \mathrm{\Gamma }(x_n{)}^{a_n}{)}^{\frac{1}{a_1+\cdots +a_n}}.}$$ 

There are also bounds on ratios of gamma functions. The best-known isGautschi's inequality, which says that for any positive real numberx and anys ∈ (0, 1), 

The behavior of${\displaystyle \mathrm{\Gamma }(x)}$  for an increasing positive real variable is given byStirling's formula$${\displaystyle \mathrm{\Gamma }(x+1)\sim \sqrt{2\pi x}{\left(\frac{x}{e}\right)}^x,}$$ where the symbol${\displaystyle \sim }$  means asymptotic convergence: the ratio of the two sides converges to 1 in the limit$x\to +\mathrm{\infty }$ .^[1] This growth is faster than exponential,${\displaystyle \exp (\beta x)}$ , for any fixed value of${\displaystyle \beta }$ .

Another useful limit for asymptotic approximations for${\displaystyle x\to +\mathrm{\infty }}$  is:$${\displaystyle \mathrm{\Gamma }(x+\alpha )\sim \mathrm{\Gamma }(x)x^{\alpha },\alpha \in \mathbb{C}.}$$ 

When writing the error term as an infinite product, Stirling's formula can be used to define the gamma function:^[12]$${\displaystyle \mathrm{\Gamma }(x)=\sqrt{\frac{2\pi }{x}}{\left(\frac{x}{e}\right)}^x\prod _{n=0}^{\mathrm{\infty }}\left[\frac{1}{e}{\left(1+\frac{1}{x+n}\right)}^{x+n+\frac{1}{2}}\right]}$$ 

Although the main definition of the gamma function—the Euler integral of the second kind—is only valid (on the real axis) for positive arguments, its domain can be extended withanalytic continuation^[13] to negative arguments by shifting the negative argument to positive values by using either the Euler's reflection formula,$${\displaystyle \mathrm{\Gamma }(-x)=\frac{1}{\mathrm{\Gamma }(x+1)}\frac{\pi }{\sin (\pi (x+1))},}$$ or the fundamental property,$${\displaystyle \mathrm{\Gamma }(-x):=\frac{1}{-x}\mathrm{\Gamma }(-x+1),}$$ when${\displaystyle x\notin \mathbb{Z}}$ . For example,$${\displaystyle \mathrm{\Gamma }\left(-\frac{1}{2}\right)=-2\mathrm{\Gamma }\left(\frac{1}{2}\right).}$$ 

The behavior for non-positive${\displaystyle z}$  is more intricate. Euler's integral does not converge for${\displaystyle \mathrm{\Re }(z)\le 0}$ , but the function it defines in the positive complex half-plane has a uniqueanalytic continuation to the negative half-plane. One way to find that analytic continuation is to use Euler's integral for positive arguments and extend the domain to negative numbers by repeated application of the recurrence formula,^[1]$${\displaystyle \mathrm{\Gamma }(z)=\frac{\mathrm{\Gamma }(z+n+1)}{z(z+1)\cdots (z+n)},}$$ choosing${\displaystyle n}$  such that${\displaystyle z+n}$  is positive. The product in the denominator is zero when${\displaystyle z}$  equals any of the integers${\displaystyle 0,-1,-2,\dots }$ . Thus, the gamma function must be undefined at those points to avoiddivision by zero; it is ameromorphic function withsimple poles at the non-positive integers.^[1]

For a function${\displaystyle f}$  of a complex variable${\displaystyle z}$ , at asimple pole${\displaystyle c}$ , theresidue of${\displaystyle f}$  is given by:$${\displaystyle \mathrm{Res}(f,c)=\underset{z\to c}{lim}(z-c)f(z).}$$ 

For the simple pole${\displaystyle z=-n}$ , the recurrence formula can be rewritten as:$${\displaystyle (z+n)\mathrm{\Gamma }(z)=\frac{\mathrm{\Gamma }(z+n+1)}{z(z+1)\cdots (z+n-1)}.}$$ The numerator at${\displaystyle z=-n,}$  is$${\displaystyle \mathrm{\Gamma }(z+n+1)=\mathrm{\Gamma }(1)=1}$$ and the denominator$${\displaystyle z(z+1)\cdots (z+n-1)=-n(1-n)\cdots (n-1-n)=(-1{)}^{n}n!.}$$ So the residues of the gamma function at those points are:^[14]$${\displaystyle \mathrm{Res}(\mathrm{\Gamma },-n)=\frac{(-1{)}^n}{n!}.}$$ The gamma function is non-zero everywhere along the real line, although it comes arbitrarily close to zero asz → −∞. There is in fact no complex number${\displaystyle z}$  for which${\displaystyle \mathrm{\Gamma }(z)=0}$ , and hence thereciprocal gamma function$\frac{1}{\mathrm{\Gamma }(z)}$  is anentire function, with zeros at${\displaystyle z=0,-1,-2,\dots }$ .^[1]

On the real line, the gamma function has a local minimum atz_min ≈+1.46163214496836234126^[15] where it attains the valueΓ(z_min) ≈+0.88560319441088870027.^[16] The gamma function rises to either side of this minimum. The solution toΓ(z − 0.5) = Γ(z + 0.5) isz = +1.5 and the common value isΓ(1) = Γ(2) = +1. The positive solution toΓ(z − 1) = Γ(z + 1) isz =φ ≈ +1.618, thegolden ratio, and the common value isΓ(φ − 1) = Γ(φ + 1) =φ! ≈+1.44922960226989660037.^[17]

The gamma function must alternate sign between its poles at the non-positive integers because the product in the forward recurrence contains an odd number of negative factors if the number of poles between${\displaystyle z}$  and${\displaystyle z+n}$  is odd, and an even number if the number of poles is even.^[14] The values at the local extrema of the gamma function along the real axis between the non-positive integers are:

Γ(−0.50408300826445540925...^[18]) =−3.54464361115500508912...,

Γ(−1.57349847316239045877...^[19]) =2.30240725833968013582...,

Γ(−2.61072086844414465000...^[20]) =−0.88813635840124192009...,

Γ(−3.63529336643690109783...^[21]) =0.24512753983436625043...,

Γ(−4.65323776174314244171...^[22]) =−0.05277963958731940076..., etc.

There are many formulas, besides the Euler integral of the second kind, that express the gamma function as an integral. For instance, when the real part ofz is positive,^[23]$${\displaystyle \mathrm{\Gamma }(z)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{zt-e^t}dt}$$ and^[24]$${\displaystyle \mathrm{\Gamma }(z)={\int }_0^1{\left(\log \frac{1}{t}\right)}^{z-1}dt,}$$ $${\displaystyle \mathrm{\Gamma }(z)=2c^z{\int }_0^{\mathrm{\infty }}{t}^{2z-1}{e}^{-c{t}^2}dt,c>0}$$ where the three integrals respectively follow from the substitutions${\displaystyle t=e^{-x}}$ ,${\displaystyle t=-\log x}$ ^[25] and${\displaystyle t=c{x}^2}$ ^[26] in Euler's second integral. The last integral in particular makes clear the connection between the gamma function at half integer arguments and theGaussian integral: if${\displaystyle z=1/2,c=1}$  we get$${\displaystyle \mathrm{\Gamma }(1/2)=2{\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt=\sqrt{\pi }.}$$ 

Binet's first integral formula for the gamma function states that, when the real part ofz is positive, then:^[27]$${\displaystyle \mathrm{l}\mathrm{o}\mathrm{g}\mathrm{\Gamma }(z)=\left(z-\frac{1}{2}\right)\log z-z+\frac{1}{2}\log (2\pi )+{\int }_0^{\mathrm{\infty }}\left(\frac{1}{2}-\frac{1}{t}+\frac{1}{e^t-1}\right)\frac{e^{-tz}}{t}dt.}$$ The integral on the right-hand side may be interpreted as aLaplace transform. That is,$${\displaystyle \log \left(\mathrm{\Gamma }(z){\left(\frac{e}{z}\right)}^z\sqrt{\frac{z}{2\pi }}\right)=\mathcal{L}\left(\frac{1}{2t}-\frac{1}{t^2}+\frac{1}{t(e^t-1)}\right)(z).}$$ 

Binet's second integral formula states that, again when the real part ofz is positive, then:^[28]$${\displaystyle \mathrm{l}\mathrm{o}\mathrm{g}\mathrm{\Gamma }(z)=\left(z-\frac{1}{2}\right)\log z-z+\frac{1}{2}\log (2\pi )+2{\int }_0^{\mathrm{\infty }}\frac{\arctan (t/z)}{e^{2\pi t}-1}dt.}$$ 

LetC be aHankel contour, meaning a path that begins and ends at the point∞ on theRiemann sphere, whose unit tangent vector converges to−1 at the start of the path and to1 at the end, which haswinding number 1 around0, and which does not cross[0, ∞). Fix a branch of${\displaystyle \log (-t)}$  by taking a branch cut along[0, ∞) and by taking${\displaystyle \log (-t)}$  to be real whent is on the negative real axis. Assumez is not an integer. Then Hankel's formula for the gamma function is:^[29]$${\displaystyle \mathrm{\Gamma }(z)=-\frac{1}{2i\sin \pi z}{\int }_C(-t{)}^{z-1}{e}^{-t}dt,}$$ where${\displaystyle (-t{)}^{z-1}}$  is interpreted as${\displaystyle \exp ((z-1)\log (-t))}$ . The reflection formula leads to the closely related expression$${\displaystyle \frac{1}{\mathrm{\Gamma }(z)}=\frac{i}{2\pi }{\int }_C(-t{)}^{-z}{e}^{-t}dt,}$$ again valid wheneverz is not an integer.

The gamma function can also be represented by a sum of twocontinued fractions:^[30][31] where${\displaystyle z\in \mathbb{C}}$ .