Could you explain how to do this in detail? As you know, given z E C, z + 0, then z/2 is a set containing two distinct elements, so thatthe formula f(z) = z'/2 does not define a function. A way to define the complex squareroot 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 ofits argument (so that Arg(z) E (-1, 7]). Define a function f(2) by settingf(2) =Vizlei Arg(z)/2 if z + 0,f(0) = 0,and then define a function g(z) by settingg(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)²have the right to be called complex square root. The fact that there is more than onecomplex 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 functionp(t) = f(e*)of the real variable t e (-n, T] with complex values. Show thatlim e(t) = -i,t→(-x)+lim p(t) = i .t→TーDo you think thatlim f(z)z--1exists? Why?

Name: Could you explain how to do this in detail? As you know, given z E C,
Uploaded: 2024-03-20T06:38:44.000Z
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Description: Could you explain how to do this in detail? As you know, given z E C, z + 0, then z/2 is a set containing two distinct elements, so thatthe formula f(z) = z'/2 does not define a function. A way to define the complex squareroot as a function (actually

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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?