# §15.10 Hypergeometric Differential Equation

## §15.10(i) Fundamental Solutions

 15.10.1 $z(1-z)\frac{{\mathrm{d}}^{2}w}{{\mathrm{d}z}^{2}}+\left(c-(a+b+1)z\right)\frac% {\mathrm{d}w}{\mathrm{d}z}-abw=0.$

This is the hypergeometric differential equation. It has regular singularities at $z=0,1,\infty$, with corresponding exponent pairs $\{0,1-c\}$, $\{0,c-a-b\}$, $\{a,b\}$, respectively. When none of the exponent pairs differ by an integer, that is, when none of $c$, $c-a-b$, $a-b$ is an integer, we have the following pairs $f_{1}(z)$, $f_{2}(z)$ of fundamental solutions. They are also numerically satisfactory (§2.7(iv)) in the neighborhood of the corresponding singularity.

### Singularity $z=0$

 15.10.2 $\displaystyle f_{1}(z)$ $\displaystyle=F\left({a,b\atop c};z\right),$ $\displaystyle f_{2}(z)$ $\displaystyle=z^{1-c}F\left({a-c+1,b-c+1\atop 2-c};z\right),$
 15.10.3 $\mathscr{W}\left\{f_{1}(z),f_{2}(z)\right\}=(1-c)z^{-c}(1-z)^{c-a-b-1}.$

### Singularity $z=1$

 15.10.4 $\displaystyle f_{1}(z)$ $\displaystyle=F\left({a,b\atop a+b+1-c};1-z\right),$ $\displaystyle f_{2}(z)$ $\displaystyle=(1-z)^{c-a-b}F\left({c-a,c-b\atop c-a-b+1};1-z\right),$
 15.10.5 $\mathscr{W}\left\{f_{1}(z),f_{2}(z)\right\}=(a+b-c)z^{-c}(1-z)^{c-a-b-1}.$

### Singularity $z=\infty$

 15.10.6 $\displaystyle f_{1}(z)$ $\displaystyle=z^{-a}F\left({a,a-c+1\atop a-b+1};\frac{1}{z}\right),$ $\displaystyle f_{2}(z)$ $\displaystyle=z^{-b}F\left({b,b-c+1\atop b-a+1};\frac{1}{z}\right),$
 15.10.7 $\mathscr{W}\left\{f_{1}(z),f_{2}(z)\right\}=(a-b)z^{-c}(z-1)^{c-a-b-1}.$

(a) If $c$ equals $n=1,2,3,\dots$, and $a=1,2,\dots,n-1$, then fundamental solutions in the neighborhood of $z=0$ are given by (15.10.2) with the interpretation (15.2.5) for $f_{2}(z)$.

(b) If $c$ equals $n=1,2,3,\dots$, and $a\neq 1,2,\dots,n-1$, then fundamental solutions in the neighborhood of $z=0$ are given by $F\left(a,b;n;z\right)$ and

 15.10.8 $F\left({a,b\atop n};z\right)\ln z-\sum_{k=1}^{n-1}\frac{(n-1)!(k-1)!}{(n-k-1)!% {\left(1-a\right)_{k}}{\left(1-b\right)_{k}}}(-z)^{-k}\\ +\sum_{k=0}^{\infty}\frac{{\left(a\right)_{k}}{\left(b\right)_{k}}}{{\left(n% \right)_{k}}k!}z^{k}\left(\psi\left(a+k\right)+\psi\left(b+k\right)-\psi\left(% 1+k\right)-\psi\left(n+k\right)\right),$ $a,b\neq n-1,n-2,\dots,0,-1,-2,\dots$, ⓘ Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, ${\left(\NVar{a}\right)_{\NVar{n}}}$: Pochhammer’s symbol (or shifted factorial), $\psi\left(\NVar{z}\right)$: psi (or digamma) function, $!$: factorial (as in $n!$), $\ln\NVar{z}$: principal branch of logarithm function, $n$: integer, $z$: complex variable, $a$: real or complex parameter, $b$: real or complex parameter, $k$: integer and $m$: integer A&S Ref: 15.5.19. (Compared with (15.10.8) this result has a multiple of $F\left(a,b;n;z\right)$ added to the right-hand side. It also contains an error: the conditions on $a$ and $b$ should be $a,b\neq 1,2,\dots,m$.) Referenced by: (15.10.8) Permalink: http://dlmf.nist.gov/15.10.E8 Encodings: TeX, pMML, png See also: Annotations for §15.10(i), §15.10(i), §15.10 and Ch.15

or

 15.10.9 $F\left({-m,b\atop n};z\right)\ln z-\sum_{k=1}^{n-1}\frac{(n-1)!(k-1)!}{(n-k-1)% !{\left(m+1\right)_{k}}{\left(1-b\right)_{k}}}(-z)^{-k}+\sum_{k=0}^{m}\frac{{% \left(-m\right)_{k}}{\left(b\right)_{k}}}{{\left(n\right)_{k}}k!}z^{k}\left(% \psi\left(1+m-k\right)+\psi\left(b+k\right)-\psi\left(1+k\right)-\psi\left(n+k% \right)\right)+(-1)^{m}m!\sum_{k=m+1}^{\infty}\frac{(k-1-m)!{\left(b\right)_{k% }}}{{\left(n\right)_{k}}k!}z^{k},$ $a=-m$, $m=0,1,2,\dots$; $b\neq n-1,n-2,\dots,0,-1,-2,\dots$,

or

 15.10.10 $F\left({-m,-\ell\atop n};z\right)\ln z-\sum_{k=1}^{n-1}\frac{(n-1)!(k-1)!}{(n-% k-1)!{\left(m+1\right)_{k}}{\left(\ell+1\right)_{k}}}(-z)^{-k}+\sum_{k=0}^{% \ell}\frac{{\left(-m\right)_{k}}{\left(-\ell\right)_{k}}}{{\left(n\right)_{k}}% k!}z^{k}\left(\psi\left(1+m-k\right)+\psi\left(1+\ell-k\right)-\psi\left(1+k% \right)-\psi\left(n+k\right)\right)+(-1)^{\ell}\ell\,!\sum_{k=\ell+1}^{m}\frac% {(k-1-\ell)!{\left(-m\right)_{k}}}{{\left(n\right)_{k}}k!}z^{k},$ $a=-m$, $m=0,1,2,\dots$; $b=-\ell$, $\ell=0,1,2,\dots,m$.

Moreover, in (15.10.9) and (15.10.10) the symbols $a$ and $b$ are interchangeable.

(c) If the parameter $c$ in the differential equation equals $2-n=0,-1,-2,\dots$, then fundamental solutions in the neighborhood of $z=0$ are given by $z^{n-1}$ times those in (a) and (b), with $a$ and $b$ replaced throughout by $a+n-1$ and $b+n-1$, respectively.

(d) If $a+b+1-c$ equals $n=1,2,3,\dots$, or $2-n=0,-1,-2,\dots$, then fundamental solutions in the neighborhood of $z=1$ are given by those in (a), (b), and (c) with $z$ replaced by $1-z$.

(e) Finally, if $a-b+1$ equals $n=1,2,3,\dots$, or $2-n=0,-1,-2,\dots$, then fundamental solutions in the neighborhood of $z=\infty$ are given by $z^{-a}$ times those in (a), (b), and (c) with $b$ and $z$ replaced by $a-c+1$ and $\ifrac{1}{z}$, respectively.

## §15.10(ii) Kummer’s 24 Solutions and Connection Formulas

The three pairs of fundamental solutions given by (15.10.2), (15.10.4), and (15.10.6) can be transformed into 18 other solutions by means of (15.8.1), leading to a total of 24 solutions known as Kummer’s solutions.

 15.10.11 $\displaystyle w_{1}(z)$ $\displaystyle=F\left({a,b\atop c};z\right)=(1-z)^{c-a-b}F\left({c-a,c-b\atop c% };z\right)=(1-z)^{-a}F\left({a,c-b\atop c};\frac{z}{z-1}\right)=(1-z)^{-b}F% \left({c-a,b\atop c};\frac{z}{z-1}\right).$ ⓘ Defines: $w_{1}(z)$: Kummer’s solution $w_{1}$ (locally) Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $z$: complex variable, $a$: real or complex parameter, $b$: real or complex parameter and $c$: real or complex parameter Permalink: http://dlmf.nist.gov/15.10.E11 Encodings: TeX, pMML, png See also: Annotations for §15.10(ii), §15.10 and Ch.15 15.10.12 $\displaystyle w_{2}(z)$ $\displaystyle={z^{1-c}}F\left({a-c+1,b-c+1\atop 2-c};z\right)$ $\displaystyle={z^{1-c}(1-z)^{c-a-b}}\*F\left({1-a,1-b\atop 2-c};z\right)$ $\displaystyle={z^{1-c}(1-z)^{c-a-1}}\*F\left({a-c+1,1-b\atop 2-c};\frac{z}{z-1% }\right)$ $\displaystyle={z^{1-c}(1-z)^{c-b-1}}\*F\left({1-a,b-c+1\atop 2-c};\frac{z}{z-1% }\right).$ ⓘ Defines: $w_{2}(z)$: Kummer’s solution $w_{2}$ (locally) Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $z$: complex variable, $a$: real or complex parameter, $b$: real or complex parameter and $c$: real or complex parameter Permalink: http://dlmf.nist.gov/15.10.E12 Encodings: TeX, pMML, png See also: Annotations for §15.10(ii), §15.10 and Ch.15 15.10.13 $\displaystyle w_{3}(z)$ $\displaystyle=F\left({a,b\atop a+b-c+1};1-z\right)$ $\displaystyle=z^{1-c}F\left({a-c+1,b-c+1\atop a+b-c+1};1-z\right)$ $\displaystyle=z^{-a}F\left({a,a-c+1\atop a+b-c+1};1-\frac{1}{z}\right)$ $\displaystyle=z^{-b}F\left({b,b-c+1\atop a+b-c+1};1-\frac{1}{z}\right).$ ⓘ Defines: $w_{3}(z)$: Kummer’s solution $w_{3}$ (locally) Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $z$: complex variable, $a$: real or complex parameter, $b$: real or complex parameter and $c$: real or complex parameter Permalink: http://dlmf.nist.gov/15.10.E13 Encodings: TeX, pMML, png See also: Annotations for §15.10(ii), §15.10 and Ch.15 15.10.14 $\displaystyle w_{4}(z)$ $\displaystyle=(1-z)^{c-a-b}F\left({c-a,c-b\atop c-a-b+1};1-z\right)$ $\displaystyle=z^{1-c}(1-z)^{c-a-b}F\left({1-a,1-b\atop c-a-b+1};1-z\right)$ $\displaystyle=z^{a-c}(1-z)^{c-a-b}F\left({1-a,c-a\atop c-a-b+1};1-\frac{1}{z}\right)$ $\displaystyle=z^{b-c}(1-z)^{c-a-b}F\left({1-b,c-b\atop c-a-b+1};1-\frac{1}{z}% \right).$ ⓘ Defines: $w_{4}(z)$: Kummer’s solution $w_{4}$ (locally) Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $z$: complex variable, $a$: real or complex parameter, $b$: real or complex parameter and $c$: real or complex parameter Permalink: http://dlmf.nist.gov/15.10.E14 Encodings: TeX, pMML, png See also: Annotations for §15.10(ii), §15.10 and Ch.15 15.10.15 $\displaystyle w_{5}(z)$ $\displaystyle={\mathrm{e}}^{a\pi\mathrm{i}}z^{-a}\*F\left({a,a-c+1\atop a-b+1}% ;\frac{1}{z}\right)$ $\displaystyle={\mathrm{e}}^{(c-b)\pi\mathrm{i}}z^{b-c}(1-z)^{c-a-b}\*F\left({1% -b,c-b\atop a-b+1};\frac{1}{z}\right)$ $\displaystyle=(1-z)^{-a}F\left({a,c-b\atop a-b+1};\frac{1}{1-z}\right)$ $\displaystyle={\mathrm{e}}^{(c-1)\pi\mathrm{i}}z^{1-c}(1-z)^{c-a-1}\*F\left({1% -b,a-c+1\atop a-b+1};\frac{1}{1-z}\right).$ ⓘ Defines: $w_{5}(z)$: Kummer’s solution $w_{5}$ (locally) Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit, $z$: complex variable, $a$: real or complex parameter, $b$: real or complex parameter and $c$: real or complex parameter Permalink: http://dlmf.nist.gov/15.10.E15 Encodings: TeX, pMML, png See also: Annotations for §15.10(ii), §15.10 and Ch.15 15.10.16 $\displaystyle w_{6}(z)$ $\displaystyle={\mathrm{e}}^{b\pi\mathrm{i}}z^{-b}F\left({b,b-c+1\atop b-a+1};% \frac{1}{z}\right)$ $\displaystyle={\mathrm{e}}^{(c-a)\pi\mathrm{i}}z^{a-c}(1-z)^{c-a-b}\*F\left({1% -a,c-a\atop b-a+1};\frac{1}{z}\right)$ $\displaystyle=(1-z)^{-b}F\left({b,c-a\atop b-a+1};\frac{1}{1-z}\right)$ $\displaystyle={\mathrm{e}}^{(c-1)\pi\mathrm{i}}z^{1-c}(1-z)^{c-b-1}\*F\left({1% -a,b-c+1\atop b-a+1};\frac{1}{1-z}\right).$ ⓘ Defines: $w_{6}(z)$: Kummer’s solution $w_{6}$ (locally) Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit, $z$: complex variable, $a$: real or complex parameter, $b$: real or complex parameter and $c$: real or complex parameter Permalink: http://dlmf.nist.gov/15.10.E16 Encodings: TeX, pMML, png See also: Annotations for §15.10(ii), §15.10 and Ch.15

The $\genfrac{(}{)}{0.0pt}{}{6}{3}=20$ connection formulas for the principal branches of Kummer’s solutions are:

 15.10.17 $\displaystyle w_{3}(z)$ $\displaystyle=\frac{\Gamma\left(1-c\right)\Gamma\left(a+b-c+1\right)}{\Gamma% \left(a-c+1\right)\Gamma\left(b-c+1\right)}w_{1}(z)+\frac{\Gamma\left(c-1% \right)\Gamma\left(a+b-c+1\right)}{\Gamma\left(a\right)\Gamma\left(b\right)}w_% {2}(z),$ 15.10.18 $\displaystyle w_{4}(z)$ $\displaystyle=\frac{\Gamma\left(1-c\right)\Gamma\left(c-a-b+1\right)}{\Gamma% \left(1-a\right)\Gamma\left(1-b\right)}w_{1}(z)+\frac{\Gamma\left(c-1\right)% \Gamma\left(c-a-b+1\right)}{\Gamma\left(c-a\right)\Gamma\left(c-b\right)}w_{2}% (z),$ 15.10.19 $\displaystyle w_{5}(z)$ $\displaystyle=\frac{\Gamma\left(1-c\right)\Gamma\left(a-b+1\right)}{\Gamma% \left(a-c+1\right)\Gamma\left(1-b\right)}w_{1}(z)+{\mathrm{e}}^{(c-1)\pi% \mathrm{i}}\frac{\Gamma\left(c-1\right)\Gamma\left(a-b+1\right)}{\Gamma\left(a% \right)\Gamma\left(c-b\right)}w_{2}(z),$ 15.10.20 $\displaystyle w_{6}(z)$ $\displaystyle=\frac{\Gamma\left(1-c\right)\Gamma\left(b-a+1\right)}{\Gamma% \left(b-c+1\right)\Gamma\left(1-a\right)}w_{1}(z)+{\mathrm{e}}^{(c-1)\pi% \mathrm{i}}\frac{\Gamma\left(c-1\right)\Gamma\left(b-a+1\right)}{\Gamma\left(b% \right)\Gamma\left(c-a\right)}w_{2}(z).$
 15.10.21 $\displaystyle w_{1}(z)$ $\displaystyle=\frac{\Gamma\left(c\right)\Gamma\left(c-a-b\right)}{\Gamma\left(% c-a\right)\Gamma\left(c-b\right)}w_{3}(z)+\frac{\Gamma\left(c\right)\Gamma% \left(a+b-c\right)}{\Gamma\left(a\right)\Gamma\left(b\right)}w_{4}(z),$ 15.10.22 $\displaystyle w_{2}(z)$ $\displaystyle=\frac{\Gamma\left(2-c\right)\Gamma\left(c-a-b\right)}{\Gamma% \left(1-a\right)\Gamma\left(1-b\right)}w_{3}(z)+\frac{\Gamma\left(2-c\right)% \Gamma\left(a+b-c\right)}{\Gamma\left(a-c+1\right)\Gamma\left(b-c+1\right)}w_{% 4}(z),$ 15.10.23 $\displaystyle w_{5}(z)$ $\displaystyle={\mathrm{e}}^{a\pi\mathrm{i}}\frac{\Gamma\left(a-b+1\right)% \Gamma\left(c-a-b\right)}{\Gamma\left(1-b\right)\Gamma\left(c-b\right)}w_{3}(z% )+{\mathrm{e}}^{(c-b)\pi\mathrm{i}}\frac{\Gamma\left(a-b+1\right)\Gamma\left(a% +b-c\right)}{\Gamma\left(a\right)\Gamma\left(a-c+1\right)}w_{4}(z),$ 15.10.24 $\displaystyle w_{6}(z)$ $\displaystyle={\mathrm{e}}^{b\pi\mathrm{i}}\frac{\Gamma\left(b-a+1\right)% \Gamma\left(c-a-b\right)}{\Gamma\left(1-a\right)\Gamma\left(c-a\right)}w_{3}(z% )+{\mathrm{e}}^{(c-a)\pi\mathrm{i}}\frac{\Gamma\left(b-a+1\right)\Gamma\left(a% +b-c\right)}{\Gamma\left(b\right)\Gamma\left(b-c+1\right)}w_{4}(z).$
 15.10.25 $\displaystyle w_{1}(z)$ $\displaystyle=\frac{\Gamma\left(c\right)\Gamma\left(b-a\right)}{\Gamma\left(b% \right)\Gamma\left(c-a\right)}w_{5}(z)+\frac{\Gamma\left(c\right)\Gamma\left(a% -b\right)}{\Gamma\left(a\right)\Gamma\left(c-b\right)}w_{6}(z),$ 15.10.26 $\displaystyle w_{2}(z)$ $\displaystyle={\mathrm{e}}^{(1-c)\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)% \Gamma\left(b-a\right)}{\Gamma\left(1-a\right)\Gamma\left(b-c+1\right)}w_{5}(z% )+{\mathrm{e}}^{(1-c)\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)\Gamma\left(a-b% \right)}{\Gamma\left(1-b\right)\Gamma\left(a-c+1\right)}w_{6}(z),$ 15.10.27 $\displaystyle w_{3}(z)$ $\displaystyle={\mathrm{e}}^{-a\pi\mathrm{i}}\frac{\Gamma\left(a+b-c+1\right)% \Gamma\left(b-a\right)}{\Gamma\left(b\right)\Gamma\left(b-c+1\right)}w_{5}(z)+% {\mathrm{e}}^{-b\pi\mathrm{i}}\frac{\Gamma\left(a+b-c+1\right)\Gamma\left(a-b% \right)}{\Gamma\left(a\right)\Gamma\left(a-c+1\right)}w_{6}(z),$ 15.10.28 $\displaystyle w_{4}(z)$ $\displaystyle={\mathrm{e}}^{(b-c)\pi\mathrm{i}}\frac{\Gamma\left(c-a-b+1\right% )\Gamma\left(b-a\right)}{\Gamma\left(1-a\right)\Gamma\left(c-a\right)}w_{5}(z)% +{\mathrm{e}}^{(a-c)\pi\mathrm{i}}\frac{\Gamma\left(c-a-b+1\right)\Gamma\left(% a-b\right)}{\Gamma\left(1-b\right)\Gamma\left(c-b\right)}w_{6}(z).$
 15.10.29 $\displaystyle w_{1}(z)$ $\displaystyle={\mathrm{e}}^{b\pi\mathrm{i}}\frac{\Gamma\left(c\right)\Gamma% \left(a-c+1\right)}{\Gamma\left(a+b-c+1\right)\Gamma\left(c-b\right)}w_{3}(z)+% {\mathrm{e}}^{(b-c)\pi\mathrm{i}}\frac{\Gamma\left(c\right)\Gamma\left(a-c+1% \right)}{\Gamma\left(b\right)\Gamma\left(a-b+1\right)}w_{5}(z),$ 15.10.30 $\displaystyle w_{1}(z)$ $\displaystyle={\mathrm{e}}^{a\pi\mathrm{i}}\frac{\Gamma\left(c\right)\Gamma% \left(b-c+1\right)}{\Gamma\left(a+b-c+1\right)\Gamma\left(c-a\right)}w_{3}(z)+% {\mathrm{e}}^{(a-c)\pi\mathrm{i}}\frac{\Gamma\left(c\right)\Gamma\left(b-c+1% \right)}{\Gamma\left(a\right)\Gamma\left(b-a+1\right)}w_{6}(z),$ 15.10.31 $\displaystyle w_{2}(z)$ $\displaystyle={\mathrm{e}}^{(b-c+1)\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)% \Gamma\left(a\right)}{\Gamma\left(a+b-c+1\right)\Gamma\left(1-b\right)}w_{3}(z% )+{\mathrm{e}}^{(b-c)\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)\Gamma\left(a% \right)}{\Gamma\left(a-b+1\right)\Gamma\left(b-c+1\right)}w_{5}(z),$ 15.10.32 $\displaystyle w_{2}(z)$ $\displaystyle={\mathrm{e}}^{(a-c+1)\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)% \Gamma\left(b\right)}{\Gamma\left(a+b-c+1\right)\Gamma\left(1-a\right)}w_{3}(z% )+{\mathrm{e}}^{(a-c)\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)\Gamma\left(b% \right)}{\Gamma\left(b-a+1\right)\Gamma\left(a-c+1\right)}w_{6}(z).$
 15.10.33 $\displaystyle w_{1}(z)$ $\displaystyle={\mathrm{e}}^{(c-a)\pi\mathrm{i}}\frac{\Gamma\left(c\right)% \Gamma\left(1-b\right)}{\Gamma\left(a\right)\Gamma\left(c-a-b+1\right)}w_{4}(z% )+{\mathrm{e}}^{-a\pi\mathrm{i}}\frac{\Gamma\left(c\right)\Gamma\left(1-b% \right)}{\Gamma\left(a-b+1\right)\Gamma\left(c-a\right)}w_{5}(z),$ 15.10.34 $\displaystyle w_{1}(z)$ $\displaystyle={\mathrm{e}}^{(c-b)\pi\mathrm{i}}\frac{\Gamma\left(c\right)% \Gamma\left(1-a\right)}{\Gamma\left(b\right)\Gamma\left(c-a-b+1\right)}w_{4}(z% )+{\mathrm{e}}^{-b\pi\mathrm{i}}\frac{\Gamma\left(c\right)\Gamma\left(1-a% \right)}{\Gamma\left(b-a+1\right)\Gamma\left(c-b\right)}w_{6}(z),$ 15.10.35 $\displaystyle w_{2}(z)$ $\displaystyle={\mathrm{e}}^{(1-a)\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)% \Gamma\left(c-b\right)}{\Gamma\left(a-c+1\right)\Gamma\left(c-a-b+1\right)}w_{% 4}(z)+\exp{\mathrm{e}}^{-a\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)\Gamma% \left(c-b\right)}{\Gamma\left(a-b+1\right)\Gamma\left(1-a\right)}w_{5}(z),$ 15.10.36 $\displaystyle w_{2}(z)$ $\displaystyle={\mathrm{e}}^{(1-b)\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)% \Gamma\left(c-a\right)}{\Gamma\left(b-c+1\right)\Gamma\left(c-a-b+1\right)}w_{% 4}(z)+{\mathrm{e}}^{-b\pi\mathrm{i}}\frac{\Gamma\left(2-c\right)\Gamma\left(c-% a\right)}{\Gamma\left(b-a+1\right)\Gamma\left(1-b\right)}w_{6}(z).$