This text provides an accessible, self-contained and rigorous introduction to complex analysis and differential equations. Topics covered include holomorphic functions, Fourier series, ordinary and partial differential equations.

The text is divided into two parts: part one focuses on complex analysis and part two on differential equations. Each part can be read independently, so in essence this text offers two books in one. In the second part of the book, some emphasis is given to the application of complex analysis to differential equations. Half of the book consists of approximately 200 worked out problems, carefully prepared for each part of theory, plus 200 exercises of variable levels of difficulty.

Tailored to any course giving the first introduction to complex analysis or differential equations, this text assumes only a basic knowledge of linear algebra and differential and integral calculus. Moreover, the large number of examples, worked out problems and exercises makes this the ideal book for independent study.

3.25.Verify that for |z|<1. 3.26.Compute: (a) for |z|<1; (b) for |z|<1. 3.27.Verify that for each r<1 the series , with p n ∈{−1,1} for each n∈ℕ, is uniformly convergent on the set {z∈ℂ:|z|

the series in (4.1) is divergent. Finally, proceeding as in the first property, if ρ=(1/R+ε)r<1, then for every m≥p. This shows that the convergence is uniform. □ Example 4.9 Let us consider the power series . Its radius of convergence is Therefore, the series is absolutely convergent for |z|<1, divergent for |z|>1, and uniformly convergent on each ball with r<1. For points with |z|=1 one can have convergence or divergence. For example, for z=1 the series is divergent while for z=−1 it is

the exponential of a real number x coincides with the exponential of x when this is seen as a complex number. Example 1.26 For z=iπ, we have We also describe several properties of the exponential. Proposition 1.27 For every z,w∈ℂ and k∈ℤ, we have: 1. e z+w =e z e w and 1/e z =e −z ; 2. ; 3.(e z ) k =e kz ; 4. e z+i2kπ =e z . Proof Given z=x+iy and w=x′+iy′, we have In particular, and thus 1/e z =e −z . This establishes the first property. For the second, we note that The third

the cosine is not a bounded function in ℂ, in contrast to what happens in ℝ. One can show in a similar manner that the sine is also unbounded in ℂ. Example 1.31 Let us solve the equation cosz=1, that is, For w=e iz , we have 1/w=e −iz , and thus, that is, w 2−2w+1=0. This yields w=1, which is the same as e iz =1. Writing z=x+iy, with x,y∈ℝ, we obtain and it follows from e iz =1+i0 that Since e −y ≠0, we obtain sinx=0. Together with the identity cos2 x+sin2 x=1, this yields cosx=±1. But

function in a bounded interval, that is, for some constants c j ∈ℝ, where the intervals I j form a partition of the interval I, and where Setting I j =[a j ,b j ], we obtain (6.47) when t→+∞. Now let h be an absolutely Riemann-integrable function in a bounded interval I. Given δ>0, there exists a step function g:I→ℝ such that We have Since g is a step function, it follows from (6.47) that and thus, for any is sufficiently large t. Since δ is arbitrary, we conclude that (6.48)