8. Use the method in Example 2, Sec. 19, to show that f'(z) does not exist at any point z when (a) f(z) = Rez; (b) f(z) = Im z.

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Chapter2: Second-order Linear Odes
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8. Use the method in Example 2, Sec. 19, to show that f'(z) does not exist at any point
z when
(a) f(z) = Rez;
(b) f(z) = Im z.
Transcribed Image Text:8. Use the method in Example 2, Sec. 19, to show that f'(z) does not exist at any point z when (a) f(z) = Rez; (b) f(z) = Im z.
ot
2
:)
SEC. 19
(0, 0) horizontally through the points (Δx, 0) on the real axis (Fig. 28),
Δz = Δx + 10 = Δx - i0 = Δx + i0 = Δz.
In that case, expression (4) tells us that
Δη
Δω
Δε Δε
Hence if the limit of Aw/Az exists, its value must be unity. However, when Az
approaches (0, 0) vertically through the points (0, Ay) on the imaginary axis, so that
Δz = 0 + i Δy = 0 - i4y = -(0 + ίΔy) = − Δz,
we find from expression (4) that
Δy]
(0, Ay)•
Δω -Δz
Δε | Δε
Hence the limit must be -1 if it exists. Since limits are unique (Sec. 15), it follows
that dw/dz does not exist anywhere.
(0, 0)
(Δx, 0)
= 1.
= -1.
FIGURE 28
EXAMPLE 3. Consider the real-valued function f(z) = |z|2. Here
Δω
|z+Az|² - 12/²
Δε
(z + Δz)(z + Δz) – 27
Δε
Δε
and since z + Az = z+Az, this becomes
(5)
respectively, we have the expressions
Δω
ΔΕ
=ā + Δz + z
DERIVATIVES 57
Δω
Δε
Proceeding as in Example 2, where horizontal and vertical approaches of Az toward
the origin gave us
Δε
Δε
Δz = Δz and Δz = -Δz,
=ā+Δz+z when Δz = (Δ.x, 0)
Transcribed Image Text:ot 2 :) SEC. 19 (0, 0) horizontally through the points (Δx, 0) on the real axis (Fig. 28), Δz = Δx + 10 = Δx - i0 = Δx + i0 = Δz. In that case, expression (4) tells us that Δη Δω Δε Δε Hence if the limit of Aw/Az exists, its value must be unity. However, when Az approaches (0, 0) vertically through the points (0, Ay) on the imaginary axis, so that Δz = 0 + i Δy = 0 - i4y = -(0 + ίΔy) = − Δz, we find from expression (4) that Δy] (0, Ay)• Δω -Δz Δε | Δε Hence the limit must be -1 if it exists. Since limits are unique (Sec. 15), it follows that dw/dz does not exist anywhere. (0, 0) (Δx, 0) = 1. = -1. FIGURE 28 EXAMPLE 3. Consider the real-valued function f(z) = |z|2. Here Δω |z+Az|² - 12/² Δε (z + Δz)(z + Δz) – 27 Δε Δε and since z + Az = z+Az, this becomes (5) respectively, we have the expressions Δω ΔΕ =ā + Δz + z DERIVATIVES 57 Δω Δε Proceeding as in Example 2, where horizontal and vertical approaches of Az toward the origin gave us Δε Δε Δz = Δz and Δz = -Δz, =ā+Δz+z when Δz = (Δ.x, 0)
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