# proof of complex mean-value theorem

The function $h(t)=\mathrm{Re}\frac{f(a+t(b-a))-f(a)}{b-a}$ is a function defined on [0,1].
We have $h(0)=0$ and $h(1)=\mathrm{Re}\frac{f(b)-f(a)}{b-a}$.
By the ordinary mean-value theorem, there is a number $t$, $$, such that ${h}^{\prime}(t)=h(1)-h(0)$.
To evaluate ${h}^{\prime}(t)$, we use the assumption that $f$ is complex differentiable^{} (holomorphic). The derivative of $\frac{f(a+t(b-a))-f(a)}{b-a}$ is equal to ${f}^{\prime}(a+t(b-a))$, then ${h}^{\prime}(t)=\mathrm{Re}({f}^{\prime}(a+t(b-a)))$, so $u=a+t(b-a)$ satisfies the required equation.
The proof of the second assertion can be deduced from the result just proved by applying it to the function f multiplied by i.

Title | proof of complex mean-value theorem |
---|---|

Canonical name | ProofOfComplexMeanvalueTheorem |

Date of creation | 2013-03-22 14:34:39 |

Last modified on | 2013-03-22 14:34:39 |

Owner | Wolfgang (5320) |

Last modified by | Wolfgang (5320) |

Numerical id | 21 |

Author | Wolfgang (5320) |

Entry type | Proof |

Classification | msc 26A06 |