MATH 122A HW 8 SOLUTIONS
RAHUL SHAH
Problem 1. [§3.33.4(a)] Show that the result of Exercise 3 could have been obtained by writing
(−1 +
and first finding the square roots of −1 +
Solution. Notice that z0 = −1 +
√
√
3
3ı) 2 = [(−1 +
√ 1 3
3) 2 ]
√
3ı.
1
2
3ı = 2eı 3 . Thus (z0 ) 2 =
√ ı[ π +πn]
for n ∈ Z. Then,
2e 3
3
√
√
3
z02 = 2 2 eı[π+3πn] = 2 2eıπ eı3πn = ±2 2,
where the sign depends on the parity of the chosen n.
Problem 2. [§3.33.9] Assuming f 0 (z) exists, state the formula for the derivative of cf (z) .
Solution. Recall that
cf (z) = ef (z) log c .
Thus
d f (z)
c
dz
=
=
=
d
(f (z) log c) ef (z) log c
dz
d
(f (z) log c) cf (z)
dz
f 0 (z)(log c)cf (z) .
Problem 3. [§3.34.8] Point out how it follows from expressions (15) and (16) in §34 for | sin z|2 and | cos z|2 that
a. | sin z| ≥ | sin x|;
b. | cos z| ≥ | cos x|.
Solution. We are given that | sin z|2 = sin2 x + sinh2 y. Thus
| sin z|2
≥
sin2 x + sinh2 y
=
| sin2 x + sinh2 y| since the above number is non-negative;
≥
| sin2 x| + | sinh2 y|
≥
| sin2 x|
=
| sin x|2 .
Thus | sin z| ≥ | sin x|. The other result follows similarly.
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