WA 3: Solutions to some problems
Problem 1. Show that for any zsuch that |Im z| ≥ δ, with some δ > 0,
|tan z| ≤ 1 + 1
sinh2δ
1
2
,|cot z| ≤ 1 + 1
sinh2δ
1
2
.
Solution. We have
|tan z|2=|sin z|2
|cos z|2=sin2x+ sinh2y
cos2x+ sinh2y≤1 + sinh2y
sinh2y= 1 + 1
sinh2y.
Since sinh2yis an even function, it suffices to show that sinh2y≥sinh2δwhen
y≥δ. This follows from the fact that (sinh2y)0= 2 sinh ycosh y > 0 when y > 0,
and thus sinh2yis increasing for positive y. Therefore,
|tan z|2≤1 + 1
sinh2y≤1 + 1
sinh2δ.
The second inequality is proved analogously.
Problem 2. Show that
(a) the set of values of log(i1/2) is the same as the set of values of (1/2)log i;
(b) the set of values of log(i2) is not the same as the set of values of 2 logi. Does
this contradict to the identity log(z1z2) = log(z1) + log(z2)?
Solution. (b) We have log(i2) = log(−1) = iπ(1 + 2n), n∈Z; 2 log i=
2iπ
2+ 2πk=iπ(1 + 4πk), k∈Z. Thus, log(i2)6= 2 log ias two sets. This does
not contradicts to the identity log(z1z2) = log(z1) + log(z2) because
log i+ log i=iπ
2+ 2πk+iπ
2+ 2πn=iπ(1 + 2(k+n)), k , n ∈Z,
and thus log(i2) = log i+ log ias two sets.
This example shows that, in general, 2 log z6= log z+ log zas two sets.
Problem 4. Find the linear transformation w=Az +B , A 6= 0, mapping the
strip
{z=x+iy :kx +b1≤y≤kx +b2},
where k, b1, b2are real constants and b1< b2, onto the strip
{w=u+iv : 0 ≤u≤1},
so that z=ib2is mapped to w= 0.
Solution. The boundary lines are mapped to the boundary lines under the
linear transformation w=Az +B. Since z=ib2belongs to the line y=kx +b2
(for x= 0), and 0 belongs to the line u= 0, we conclude that the line y=kx +b2
is mapped to u= 0, and the line y=kx +b1is mapped to the line u= 1. The first
condition can be written as
Re (A(x+i(kx +b2)) + B) = 0,
and the second condition can be written as
Re (A(x+i(kx +b1)) + B) = 1.
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