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#Soient <math>f=(P,Q)</math> et <math>g=(R,S)</math>. Si <math>\frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}=\frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}=\frac{\partial R}{\partial x}-\frac{\partial S}{\partial y}=\frac{\partial R}{\partial y}+\frac{\partial S}{\partial x}=0</math>, alors<br><math>\left(\frac{\partial R}{\partial x}\frac{\partial P}{\partial x}+\frac{\partial R}{\partial y}\frac{\partial Q}{\partial x}\right)-\left(\frac{\partial S}{\partial x}\frac{\partial P}{\partial y}+\frac{\partial S}{\partial y}\frac{\partial Q}{\partial y}\right)=\left(\frac{\partial R}{\partial x}\frac{\partial P}{\partial x}-\frac{\partial R}{\partial y}\frac{\partial P}{\partial y}\right)-\left(-\frac{\partial R}{\partial y}\frac{\partial P}{\partial y}+\frac{\partial R}{\partial x}\frac{\partial P}{\partial x}\right)=0</math> et<br><math>\left(\frac{\partial R}{\partial x}\frac{\partial P}{\partial y}+\frac{\partial R}{\partial y}\frac{\partial Q}{\partial y}\right)+\left(\frac{\partial S}{\partial x}\frac{\partial P}{\partial x}+\frac{\partial S}{\partial y}\frac{\partial Q}{\partial x}\right)=\left(\frac{\partial R}{\partial x}\frac{\partial P}{\partial y}+\frac{\partial R}{\partial y}\frac{\partial P}{\partial x}\right)+\left(-\frac{\partial R}{\partial y}\frac{\partial P}{\partial x}-\frac{\partial R}{\partial x}\frac{\partial P}{\partial y}\right)=0</math>.
Voir aussi : [[Fonctions d'une variable complexe/Fonctions holomorphes]].
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== Exercice 11==
On considère la fonction <math>f:\R\to\C,\ x\mapsto\operatorname e^{\mathrm ix}</math>. Montrer que pour tous réels distincts <math>a</math> et <math>b</math>, il n'existe aucun réel <math>c</math> tel que <math>f(b)-f(a)=(b-a)f'(c)</math>.
{{Solution|contenu=
Pour tout réel <math>c</math>,
:<math>|(b-a)f'(c)|=\left|(b-a)\mathrm i\operatorname e^{\mathrm ic}\right|=|b-a|</math>,
alors que si <math>b\ne a</math>,
:<math>|f(b)-f(a)|=\left|\operatorname e^{\mathrm i(b+a)/2}\times2\mathrm i\sin\frac{b-a}2\right|=2\left|\sin\frac{b-a}2\right|<2\left|\frac{b-a}2\right|=|b-a|</math>.
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