1 Complex Numbers. 1.1 Sums and Products

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1 1 Complex Numbers 1.1 Sums Products Definition: The complex plane, denoted C is the set of all ordered pairs (x, y) with x, y R, where Re z = x is called the real part Imz = y is called the imaginary part. We call the x-axis the real axis. We call the y-axis the imaginary axis. Usually we write z = (x, y) or z = x + iy, this is called rectangular form. For two complex numbers, z 1 = (x 1, y 1 ) z = (x, y ) we have z 1 = z if only if x 1 = x y 1 = y For z 1 = (x 1, y 1 ) z = (x, y ), we define addition as z 1 + z = (x 1, y 1 ) + (x, y ) = (x 1 + x, y 1 + y ) multiplication as z 1 z = (x 1, y 1 )(x, y ) = (x 1 x y 1 y, y 1 x + x 1 y ) How do the real numbers interact here? For any r R we have (r, 0) C, (r, 0) + (s, 0) = (r + s, 0) For i = (0, 1) C we have (r, 0)(s, 0) = (rs, 0) i = (0, 1)(0, 1) = ( 1, 0) Now we can see that multiplication is well-defined, since z 1 z = (x 1 + iy 1 )(x + iy ) = x 1 x + i (y 1 y ) + i(y 1 x + x 1 y ) = (x 1 x y 1 y, y 1 x + x 1 y ) 1

2 1. Basic Algebraic Properties Complex numbers are associative commutative so we can rearrange drop parenthesis. z 1 + (z + z 3 ) = (z 1 + z ) + z 3 z 1 + z = z + z 1 We have an additive identity 0 = (0, 0) C, mult. identity 1 = (1, 0) C. Check these! For any z = x + iy we have an additive inverse z = x iy since x + iy + ( x iy) = 0. For z = x + iy non-zero we have a multiplicative inverse z 1 = u + iv so that zz 1 = 1 defined by (x + iy)(u + iv) = 1 so ux vy + i(uy + vx) = 1 therefore ux vy = 1 uy + vx = 0. Solving the linear equation, we get ( ) x z 1 = x + y, y x + y If z 1 z = 0 then z 1 = 0 or z = 0. To see this, suppose z 1 0, then z = z 1 = z z 1 z 1 1 = 0 z 1 1 = 0 Some interesting things happen with quotients, for example 1 i = 1 i i i = i 1 = i 4 + i (4 + i)( + 3i) = 3i ( 3i)( + 3i) = i = i14 13.

3 1.3 Vectors Moduli We can consider z = x + iy = (x, y) as a vector on the complex plane. Then the sum can also be interpreted vectorially, as follows. Definition: The modulus of a complex number z = x + iy gives the distance from z to the origin, is given by z = x + y extending the notion of absolute value for the reals, z = (Re z) + (Im z) now gives a relation between real numbers. From here z = z Re z Re z z Im z Im z z Note that z 1 < z is meaningless, but z 1 < z has meaning. Consider the difference z 1 z = (x 1 x ) + i(y 1 y ) = (x 1 x ) + (y 1 y ) What does this mean geometrically? 3

4 1. Consider What does this mean, geometrically? 3 + i = = i = = 17. The equation z = 1 is the unit circle centered at the origin. 3. The equation z 1 + 3i = is the circle with center z = (1, 3) radius. 4. Exercise: Show that z 1 z = z 1 z. 5. Exercise: Show that z = z. 1.4 Triangle Inequality Definition: The triangle inequality is given by Illustrated by figure 3. Some consequences: z 1 + z z 1 + z. We have z 1 + z z 1 z, to see this, observe hence z 1 = (z 1 + z ) + ( z ) (z 1 + z ) + z (z 1 + z ) z 1 z if z 1 z (z 1 + z ) ( z 1 z ) if z 1 < z. What does this mean geometrically? Length of one side is greater than difference of other two. 4

5 1. Suppose z is on the circle z = 1. Then z = z + ( ) z + = 1 + = 3 z = z + ( ) z = 1 = 1. This means the distance from z to is between The triangle inequality can be extended inductively, that is z 1 + z z n z 1 + z z n for any n. So for example, for z on the circle z =, we have 3 + z + z 3+ z + z = 3+ z + z Consider the degree n polynomial given by P (z) = a 0 + a 1 z + a z a n z n where n Z + a 0,..., a n C. We will show that for some positive R R, 1 P (z) < a n R n whenever z > R. Morally: Reciprocal 1/P (z) is bounded from above when z is outside of the circle z = R. Recall, z 1 z = z 1 z z = write Then Then the equation above becomes w = P (z) a nz n z n = a 0 z n + a 1 z n a n 1 z. P (z) = (a n + w)z n wz n = a 0 + a 1 z a n 1 z n 1 hence implying w z n a 0 + a 1 z a n 1 z n 1 w a 0 z n + a 1 z n a n 1 z 5

6 Now pick R large enough so that a i z n i < a n n when z > R. Then, Consequently, hence w < n a n n = a n whenever z < R. a n + w a n w > a n P (z) = a n + w z n > a n R n. 1.5 Complex Conjugates Definition: The complex conjugate of a number z = x + iy is given by z = x iy Some observations: z = z z = z. z 1 + z = z 1 + z, since z 1 + z = x 1 + iy 1 + x + iy = x i + x i(y 1 + y ) = x 1 iy 1 + x iy = z 1 + z. z 1 z = z 1 z z 1 z = z 1 z ( z1 z ) = z 1 z z + z = Re z. z z = iim z. zz = (x + iy)(x iy) = x + y = z i i = ( 1+3i)(+i) ( i)(+i) = 5+5i i = 5+5i 5 = 1 + i 6

7 . We can use conjugates to establish modulus properties. We know zz = z. Therefore, we obtain z 1 z = (z 1 z )(z 1 z ) = z 1 z z 1 z = z 1 z 1 z z = z 1 x = ( z 1 z ) from which we conclude that z 1 z = z 1 z, since modulus is always positive. 1.6 Exponential Form Let r > 0 θ be polar coordinates for z = (x, y), then then z = r(cos θ + i sin θ) z = r cos θ + sin θ = r Definition: The value θ = arg(z) is an argument for z, it is the angle z = (x, y) makes with the real axis. Infinite arguments exist for z, namely, θ + π, θ + 4π,..., θ + kπ,... but we call Θ = Arg(z) with π < Θ π the principal argument of z. 1. The complex number 1 i has Arg( 1 i) = 3π 4 but for n Z. arg( 1 i) = 5π 4 + kπ 7

8 Definition: The symbol e iθ is defined by Euler s formula as e iθ = cos θ + i sin θ hence z = re iθ = r(cosθ + i sin θ) 1. Writing 1 i in exponential form we have 1 i = e i 3π 4 or 1 i = 3π e i( +πk) 4. The set of numbers z = e iθ are the set of numbers on the unit circle. Accordingly, Also, e iπ = 1 e iπ/ = i e i4π = 1. z = Re iθ where 0 θ π is a parametrization of the unit circle, moving counter-clockwise, moves the circle to be centered at z 0. Some basic properties of exponential forms: e iθ e iθ = e i(θ+θ ) re iθ r e iθ = (rr ) e i (θ + θ ) reiθ r e iθ = r r e i(θ θ ) z = z 0 + Re iθ where 0 θ π If z = re iθ then z 1 = 1 re iθ = 1 r e iθ z n = r n e inθ. 8

9 1. From the properties above we see that arg(z 1 z ) = arg(z 1 ) + arg(z ). but this doesn t necessarily hold when arg is replaced with Arg. For example, z 1 = 1 z = i, then Arg( 1) = π Arg(i) = π so but. Find the principal argument of Arg( 1) + Arg(i) = π + π = 3π Arg( 1 i) = Arg( i) = π i 1 i. We already know that i = 1e i π 1 i = e i 3π 4 so i 1 i = ei π e i 3π 4 = 1 e i(π/+3π/4) Therefore hence ( ) i arg = π 1 i + 3π 4 = 5π 4 ( ) i Arg = 5π 1 i 4 π = 3π 4 3. Let s write the expression ( 1 + i) 7 in rectangular form. First, for z = ( 1 + i) we have so 1 + i = e i 3π 4, ( 1 + i) 7 = z = 1 + i = Arg(z) = 3π 4 ( ) e i i3π 7 i1π 4 = 8 e 4 = 8 e i5π e iπ iπ 4 = 8 e 4 But we can easily see that so (1 + i) = e iπ 4 ( 1 + i) 7 = 8(1 + i) 9

10 4. When r = 1, for any value n Z, we have ( e iθ) n = e inθ but this means this is known as De Moivre s formula. (cos θ + i sin θ) n = cos nθ + i sin nθ 5. Using the above with n =, we reclaim known identities cos θ sin θ + i cos θ sin θ = (cos θ + i sin θ) = cos θ + i sin θ. Now equating real imaginary parts, we have 1.7 Roots of Complex Numbers cos θ = cos θ sin θ sin θ = cos θ sin θ Two non-zero complex numbers can be arrived at in infinitely many ways, for example are equal precisely when z = re iθ z = r e iθ r = r θ = θ + πk for k Z. We also know that z n = r n e inθ, so we should know something about n th roots. Definition: An n th root of z 0 C is a complex number z = re iθ such that that is hence Since r 0 is real, this just means r = n r 0. For k Z, we know that z n = z 0 r 0 e iθ 0 = r n e inθ, r 0 = r n nθ = θ 0 + πk. θ = θ 0 n + πk n So the n th roots of z 0 are given by the infinite family [ ( z = n θ0 r 0 exp i n + πk )] n all lying on the circle z = n r 0. So n of them should be more special than the others, namely, those with [ ( c k = n θ0 r 0 exp i n + πk )] n k 0, 1,,..., n 1. 10

11 When z 0 is a positive real then n r 0 means the familiar unique positive root. When θ 0 = Arg(z 0 ) then we call c 0 the principal root. When z 0 is a positive real, its principal root is n r Let s find all values of ( 16) 1 4, that is, the 4 th roots of 16. Since so the roots are just 16 = 16e i(π+kπ) c k = e i( π 4 + kπ ) where k = 0, 1,, 3. So we have c k = e i( ( π 4 ) e i( kπ ) = cos π 4 + i sin π ) e i( ( kπ ) 1 = + i 1 ) e i( kπ ) = (1 + i)e i( kπ ) 4 therefore c 0 = (1 + i) 1 = (1 + i) c 1 = (1 + i) i = ( 1 + i) c = (1 + i) 1 = ( 1 i) c 3 = (1 + i) i = (1 i). Let s examine the n th roots of 1, called the n th roots of unity. We start with where k Z. Then, c k = n 1 e 1 = 1 e i(0+πk) πk i(0+ n ) = e i( πk n ) where k {0, 1,,..., n 1}. When n = we get c 0 = 1 c 1 = 1. For larger n, we get roots lying on the regular n-gon inscribed in the circle z = 1. 11

12 1.8 Regions in the Complex Plane Definition: We define an neighborhood around a given point z0 by z z0 < we can also define the deleted neighborhood, that is the neighborhood minus the point z0 itself, by 0 < z z0 < For a given set S contained in C, A point z0 is called interior if there exists a neighborhood containing only z0 points of S. A point z0 is called exterior if there exists a neighborhood containing z0 no points in S. If z0 is neither of these, then it is a boundary point. A set is open if it contains none of its boundary points. A set is closed if it contains all of its boundary points. A set is a closure of S if it contains S plus its boundary points. For example, z 1 is the closure of z < 1. Some sets like 0 < z 1 is neither open nor closed. 1

13 An open set S is connected if every pair of points in S can be joined by a polygonal line. A domain is an non-empty open connected set. A domain with some, none, or all of its boundary points is called a region. A set S is bounded if every point of S lies in some circle z < R. 1. Consider the set Im ( 1 z ) > 1 in relation to these properties, for z 0 we have 1 z = z zz = z z = x iy x + y so we want therefore Completing the square, we get Im ( ) 1 = y z x + y > 1 x + y + y < 0. which means x + y + y < 1 4 ( (x 0) + y + 1 < ) ( ) 1. This is the region whose boundary is the circle centered at x = 0 y = 1/ with radius 1/. A point z 0 is an accumulation point, or limit point, of a set S if each deleted neighborhood of z 0 contains at least one point of S. 13

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