Math 336 𝔖𝔢𝔠𝔬𝔫𝔡 𝔐𝔦𝔡𝔱𝔢𝔯𝔪 𝔖𝔬𝔩𝔲𝔱𝔦𝔬𝔫𝔰

1. Since

|(z^{7} + 5z^{3} + z − 2) − 5z^{3}| = |z^{7} + z − 2|
≤ 4
< 5 = |5z^{3}|
≤ |(z^{7} + 5z^{3} + z − 2)| + | − 5z^{3}|,

on |z| = 1, z^{7} + 5z^{3} + z − 2 has three roots in |z| < 1 and no roots on |z| = 1. Similarly on |z| = 2,

|z^{7} + 5z^{3} + z − 2 − z^{7}| ≤ 44 < 128 ≤ |z^{7} + 5z^{3} + z − 2| + |z^{7}|,

so z^{7} + 5z^{3} + z − 2 has seven roots in |z| < 2 and hence has four roots in 1 < |z| < 2.

2. By continuity, since lim_{n→∞} f(1/n) = 0, f(0) = 0. Now if f is not identically 0, then f(z) = z^{k}g(z)

where g(0) ≠ 0. So

|f(1/n)| = |\frac{g(1/n)}{n^{k}}| ≤ e^{−n},

so |g(1/n)| < n^{k}/e^{n} and g(1/n) → 0 as n → 0. This implies g(0) = 0 and is a contradiction. Hence f is identically 0.

3. f = e^{z−1/z} has a Laurent series expansion ∑_{n=−∞}^{n=+∞} a_{n}z^{n} in 0 < |z| < ∞ and the coefficients are given

by the formula

a_{n} = \frac{1}{2πi} ∫_{|z|=1} e^{z−1/z}z^{−n−1}dz.

Subsitute z = e^{it} to get

a_{n} = \frac{1}{2π} ∫_{−π}^{π} e^{e^{it}−e^{−it}−int}dt

= \frac{1}{2π} ∫_{−π}^{π} e^{i(2 sin t−nt)}dt

= \frac{1}{2π} ∫_{−π}^{π} (cos(2 sin t − nt) + i sin(2 sin t − nt)) dt.

Since sin(2 sin t − nt) is odd, its integral is 0. This proves the result.

4. We perform the following manipulations

1 + sin^{2} t = (3 − cos 2t)/2
∫_{0}^{2π} \frac{dt}{1 + sin^{2} t} = 2 ∫_{0}^{2π} \frac{dt}{3 − cos 2t}

= ∫_{0}^{4π} \frac{du}{3 − cos u}

= \frac{4}{i} ∫_{|z|=1} \frac{dz}{6z − z^{2} − 1}

= 4i ∫_{|z|=1} \frac{dz}{z^{2} − 6z + 1} = 4i ∫_{|z|=1} \frac{dz}{(z − (3 − 2\sqrt{2}))(z − (3 + 2\sqrt{2}))}

= 4i\frac{2πi}{(3 − 2\sqrt{2)} − (3 + 2\sqrt{2})}

= 4i\frac{2πi}{−4\sqrt{2}}

= \sqrt{2}π.

We have used the Cauchy integral formula for the function

\frac{1}{(z − (3 + 2\sqrt{2}))},

Which is analytic in the disk |z| < 1.