As you know, given z e C, z + 0, then z1/2 is a set containing two distinct elements, so that the formula f(z) = z1/² does not define a function. A way to define the complex square root as a function (actually, to define many complex square roots) is the following. Given z E C denote by |2| its modulus and (when z + 0) by Arg(z) the principal value of its argument (so that Arg(z) E (–1, 7]). Define a function f(z) by setting f(0) = 0, f(2) = Vlzle*Arg(=)/2 if z +0, and then define a function g(z) by setting 9(2) = -Vlz|e* Arg(=)/2 if z + 0. g(0) = 0, (i) Check that f(1) = 1 while g(1) = –1, so that f and g are different functions. = z and g(z)² = z for every z E C, so that both functions would have the right to be called complex square root. The fact that there is more than one complex square root leads to call f(2) and g(z) branches of the complex square root. (iii) The restriction of f(z) to the unit circle {eit :t e (-n, 7]} defines a function p(t) = f(et) of the real variable t e (–n, T] with complex values. Show that lim e(t) = –i, t→(-n)+ lim φ(t) = i. t→Tー Do you think that lim f(2) z--1 exists? Why?
As you know, given z e C, z + 0, then z1/2 is a set containing two distinct elements, so that the formula f(z) = z1/² does not define a function. A way to define the complex square root as a function (actually, to define many complex square roots) is the following. Given z E C denote by |2| its modulus and (when z + 0) by Arg(z) the principal value of its argument (so that Arg(z) E (–1, 7]). Define a function f(z) by setting f(0) = 0, f(2) = Vlzle*Arg(=)/2 if z +0, and then define a function g(z) by setting 9(2) = -Vlz|e* Arg(=)/2 if z + 0. g(0) = 0, (i) Check that f(1) = 1 while g(1) = –1, so that f and g are different functions. = z and g(z)² = z for every z E C, so that both functions would have the right to be called complex square root. The fact that there is more than one complex square root leads to call f(2) and g(z) branches of the complex square root. (iii) The restriction of f(z) to the unit circle {eit :t e (-n, 7]} defines a function p(t) = f(et) of the real variable t e (–n, T] with complex values. Show that lim e(t) = –i, t→(-n)+ lim φ(t) = i. t→Tー Do you think that lim f(2) z--1 exists? Why?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![As you know, given z E C, z + 0, then z/2 is a set containing two distinct elements, so that
the formula f(z) = z'/2 does not define a function. A way to define the complex square
root as a function (actually, to define many complex square roots) is the following.
Given z E C denote by |2| its modulus and (when z + 0) by Arg(z) the principal value of
its argument (so that Arg(z) E (-1, 7]). Define a function f(2) by setting
f(2) =
Vizlei Arg(z)/2 if z + 0,
f(0) = 0,
and then define a function g(z) by setting
g(0) = 0,
g(2) = -Vl2le Arg(2)/2 if z +0.
(i) Check that f(1) = 1 while g(1) = -1, so that f and g are different functions.
(ii) Check that f(z)²
have the right to be called complex square root. The fact that there is more than one
complex square root leads to call f(z) and g(z) branches of the complex square root.
= z and g(z)2
= z for every z E C, so that both functions would
(iii) The restriction of f(z) to the unit circle {et :t e (-n, T]} defines a function
p(t) = f(e*)
of the real variable t e (-n, T] with complex values. Show that
lim e(t) = -i,
t→(-x)+
lim p(t) = i .
t→Tー
Do you think that
lim f(z)
z--1
exists? Why?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2Fc5770fff-4076-4f35-924a-cfcb01c466ec%2Fn2midt_processed.png&w=3840&q=75)
Transcribed Image Text:As you know, given z E C, z + 0, then z/2 is a set containing two distinct elements, so that
the formula f(z) = z'/2 does not define a function. A way to define the complex square
root as a function (actually, to define many complex square roots) is the following.
Given z E C denote by |2| its modulus and (when z + 0) by Arg(z) the principal value of
its argument (so that Arg(z) E (-1, 7]). Define a function f(2) by setting
f(2) =
Vizlei Arg(z)/2 if z + 0,
f(0) = 0,
and then define a function g(z) by setting
g(0) = 0,
g(2) = -Vl2le Arg(2)/2 if z +0.
(i) Check that f(1) = 1 while g(1) = -1, so that f and g are different functions.
(ii) Check that f(z)²
have the right to be called complex square root. The fact that there is more than one
complex square root leads to call f(z) and g(z) branches of the complex square root.
= z and g(z)2
= z for every z E C, so that both functions would
(iii) The restriction of f(z) to the unit circle {et :t e (-n, T]} defines a function
p(t) = f(e*)
of the real variable t e (-n, T] with complex values. Show that
lim e(t) = -i,
t→(-x)+
lim p(t) = i .
t→Tー
Do you think that
lim f(z)
z--1
exists? Why?
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