By Luis Barreira

This textual content presents an obtainable, self-contained and rigorous advent to complicated research and differential equations. subject matters lined contain holomorphic capabilities, Fourier sequence, usual and partial differential equations.

The textual content is split into components: half one specializes in complicated research and half on differential equations. each one half will be learn independently, so in essence this article deals books in a single. within the moment a part of the ebook, a few emphasis is given to the appliance of complicated research to differential equations. 1/2 the e-book comprises nearly two hundred labored out difficulties, rigorously ready for every a part of thought, plus 2 hundred routines of variable degrees of difficulty.

Tailored to any path giving the 1st creation to advanced research or differential equations, this article assumes just a simple wisdom of linear algebra and differential and crucial calculus. additionally, the big variety of examples, labored out difficulties and routines makes this the right booklet for self sustaining study.

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**Sample text**

Basic Notions Solution For z = 0 we have log z = log |z| + i arg z, with arg z ∈ (−π, π]. Hence, for log z = 0 we obtain log log z = log|log z| + i arg log z = log (log |z|)2 + (arg z)2 + i arg log z, with arg log z ∈ (−π, π]. This implies that log log z is purely imaginary if and only if Re log log z = 1 log (log |z|)2 + (arg z)2 = 0, 2 which is equivalent to (log |z|)2 + (arg z)2 = 1, with z = 0. Taking α ∈ R such that log |z| = cos α and arg z = sin α, we obtain |z| = ecos α , and thus, iα z = |z|ei arg z = ecos α ei sin α = ee .

3 For the function f (z) = |z|, we have f (z) − f (z0 ) = |z| − |z0 | ≤ |z − z0 |. This implies that |f (z) − f (z0 )| < δ whenever |z − z0 | < δ, and hence, the function f is continuous in C. 4 For the function f (z) = z 2 , we have f (z) − f (z0 ) = (z − z0 )(z + z0 ) = |z − z0 | · |z − z0 + 2z0 | ≤ |z − z0 | |z − z0 | + 2|z0 | < δ(δ + 2|z0 |) whenever |z − z0 | < δ. Since δ(δ + 2|z0 |) → 0 when δ → 0, the function f is continuous in C. 5 Now we show that the function f (z) = log z is discontinuous at all points z = −x + i0 with x > 0.

6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 5−1 +i 4 √ 5+ 5 . 8 Find all complex numbers z ∈ C such that (z 2 )2 = 1. Verify that 1 + ei2x = 2eix cos x for every x ∈ R. Compute log log i. Find whether log log z can be computed for every z = 0. Find all solutions of the equation: (a) (z + 1)2 = (z − 1)2 ; (b) 2z 2 + iz + 4 = 0; (c) z 4 + z 3 + z 2 + z = 0. Solve the equation: (a) ez = 3; (b) cosh z = i; z (c) ee = 1. Solve the equation: (a) cos z sin z = 0; (b) sin z + cos z = 1; (c) sin z = sin(2z). Determine the set of points (x, y) ∈ R2 such that: (a) x + iy = |x + iy|; (b) 2|x + iy| ≤ |x + iy − 1|.