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Published byStanley Morton Modified over 9 years ago
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Mohammed Nasser Acknowledgement: Steve Cunningham
Complex Variables Mohammed Nasser Acknowledgement: Steve Cunningham
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Open Disks or Neighborhoods
Definition. The set of all points z which satisfy the inequality |z – z0|<, where is a positive real number is called an open disk or neighborhood of z0 . Remark. The unit disk, i.e., the neighborhood |z|< 1, is of particular significance. Fig 1 1
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Interior Point Definition. A point is called an interior point of S if and only if there exists at least one neighborhood of z0 which is completely contained in S. z0 S Fig 2
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Open Set. Closed Set. Definition. If every point of a set S is an interior point of S, we say that S is an open set. Definition. S is closed iff Sc is open. Theorem: S` S, i.e., S contains all of its limit points S is closed set. Sets may be neither open nor closed. Neither Open Closed Fig 3
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Connected An open set S is said to be connected if every pair of points z1 and z2 in S can be joined by a polygonal line that lies entirely in S. Roughly speaking, this means that S consists of a “single piece”, although it may contain holes. S z1 z2 Fig 4
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Domain, Region, Closure, Bounded, Compact
An open, connected set is called a domain. A region is a domain together with some, none, or all of its boundary points. The closure of a set S denoted , is the set of S together with all of its boundary. Thus A set of points S is bounded if there exists a positive real number r such that |z|<r for every z S. A region which is both closed and bounded is said to be compact.
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Open Ball It is open. Prove it. Is it connected? Yes Fig 5
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Problems The graph of |z – ( i)| < 0.05 is shown in Fig 6 It is an open set. Is it connected? Yes Fig 6
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Problems The graph of Re(z) 1 is shown in Fig 7.It is not an open set. It is a closed set. Prove it. Is it connected? Yes Fig 7
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Problems Fig 17.10 illustrates some additional open sets. Fig 8
Both are open. Prove it. Are they connected? Yes Fig 8
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Problems Both are open. Prove it. Are they connected? Yes Fig 9
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Problems It is open. Prove it. Is it connected? Yes Fig 10
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Problems The graph of |Re(z)| 1 is shown in Fig 11.It is not an open set. X= -1 Is it connected? No Fig 11
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Polar Form Polar Form Referring to Fig , we have z = r(cos + i sin ) where r = |z| is the modulus of z and is the argument of z, = arg(z). If is in the interval − < , it is called the principal argument, denoted by Arg(z).
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Fig 12
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Example Solution See Fig 13 that the point lies in the fourth quarter.
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Fig 13
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Review: Real Functions of Real Variables
Definition. Let . A function f is a rule which assigns to each element a one and only one element b , . We write f: , or in the specific case b = f(a), and call b “the image of a under f.” We call “the domain of definition of f ” or simply “the domain of f ”. We call “the range of f.” We call the set of all the images of , denoted f (), the image of the function f . We alternately call f a mapping from to .
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Real Function In effect, a function of a real variable maps from one real line to another. f Fig 14
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Complex Function Definition. Complex function of a complex variable. Let C. A function f defined on is a rule which assigns to each z a complex number w. The number w is called a value of f at z and is denoted by f(z), i.e., w = f(z). The set is called the domain of definition of f. Although the domain of definition is often a domain, it need not be.
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Remark Properties of a real-valued function of a real variable are often exhibited by the graph of the function. But when w = f(z), where z and w are complex, no such convenient graphical representation is available because each of the numbers z and w is located in a plane rather than a line. We can display some information about the function by indicating pairs of corresponding points z = (x,y) and w = (u,v). To do this, it is usually easiest to draw the z and w planes separately.
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Graph of Complex Function
y v w = f(z) x u domain of definition range z-plane w-plane Fig 15
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Example 1 Find the image of the line Re(z) = 1 under f(z) = z2.
Solution Now Re(z) = x = 1, u = 1 – y2, v = 2y.
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Fig 16
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Complex Exponential Function
Evaluate e i. Solution: You prove them.
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Periodicity
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Polar From of a Complex Number Revisited
Arithmetic Operations in Polar Form The representation of z by its real and imaginary parts is useful for addition and subtraction. For multiplication and division, representation by the polar form has apparent geometric meaning.
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Suppose we have 2 complex numbers, z1 and z2 given by :
Easier with normal form than polar form Easier with polar form than normal form magnitudes multiply! phases add!
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For a complex number z2 ≠ 0, phases subtract! magnitudes divide!
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Some Exercises (Example2)
Express and in terms of powers of and Ex. 2:
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find the nth roots of unity
k is an integer Ex. : Find the solutions to the equation Imz Rez Ex. : The three roots of are Proof:
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Ex: Solve the polynomial equation
(1) (2) (3)
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Example 3 Describe the range of the function f(z) = x2 + 2i, defined on (the domain is) the unit disk |z| 1. Solution: We have u(x,y) = x2 and v(x,y) = 2. Thus as z varies over the closed unit disk, u varies between 0 and 1, and v is constant (=2). Therefore w = f(z) = u(x,y) + iv(x,y) = x2 +2i is a line segment from w = 2i to w = 1 + 2i. v y f(z) range x u domain
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Example 4 Describe the function f(z) = z3 for z in the semidisk given by |z| 2, Im z 0. Solution: We know that the points in the sector of the semidisk from Arg z = 0 to Arg z = 2/3, when cubed cover the entire disk |w| 8 because The cubes of the remaining points of z also fall into this disk, overlapping it in the upper half-plane as depicted on the next screen.
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w = z3 y v 8 2 x u -2 2 -8 8 -8 Fig 17
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If Z is in x+iy form z3=z2..z=(x2-y2+i2xy)(x+iy)=(x3-xy2+i2x2y+ix2y-iy3 -2xy2) =(x3-3xy2) +i(3x2y-y3) If u(x,y)= (x3-3xy2) and v(x,y)=(3x2y-y3), we can write z3=f(z)=u(x,y) +iv(x,y)
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Example 5 f(z)=z2, g(z)=|z| and h(z)= D={(x,x)|x is a real number}
D={|z|<4| z is a complex number} Draw the mappings
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Logarithm Function Given a complex number z = x + iy, z 0, we define w = ln z if z = ew Let w = u + iv, then We have and also
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Logarithm of a Complex Number
DEFINITION For z 0, and = arg z, Example 6 Find the values of (a) ln (−2) (b) ln i, (c) ln (−1 – i ).
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Solution
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Example 7 Find all values of z such that Solution
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Principal Value Since Arg z is unique, there is only one value of Ln z for which z 0.
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Example 8 The principal values of example 6 are as follows.
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Important Point Each function in the collection of ln z is called a branch. The function Ln z is called the principal branch or the principal logarithm function. Some familiar properties of logarithmic function hold in complex case:
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Example 9 Suppose z1 = 1 and z2 = -1. If we take ln z1 = 2i, ln z2 = i, we get
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