Lecture 5. Complex Numbers and Euler s Formula

Transcription

1 Lecture 5. Complex Numbers and Euler s Formula University of British Columbia, Vancouver Yue-Xian Li March 017 1

2 Main purpose: To introduce some basic knowledge of complex numbers to students so that they are prepared to handle complex-valued roots when solving the characteristic polynomials for eigenvalues of a matrix. Eg: In high school, students learned that the roots of a quadratic equation ax + bx + c = 0 (a 0) are given by x 1, = b ± b 4ac a where the sign of the discriminant = b 4ac determines the following three outcomes > 0, If = 0, < 0, real roots; 1 (repeated) real root; no real root. When complex-valued roots are allowed as in the case when solving eigenvalues, however, a polynomial of degree n always has n roots (Gauss Fundamental Theorem of Algebra), of which some or all of them can be identical (repeated). Thus, a quadratic equation always has roots irrespective of the sign of.

3 5.1 Definitions and basic concepts The imaginary number i: i 1 i = 1. (1) Every imaginary number is expressed as a real-valued multiple of i: 9 = 9 1 = 9i = 3i. A complex number: z = a + bi, () where a, b are real, is the sum of a real and an imaginary number. The real part of z: Re{z} = a is a real number. The imaginary part of z: Im{z} = b is a also a real number. 3

4 A complex number represents a point (a, b) in a D space, called the complex plane. Thus, it can be regarded as a D vector expressed in form of a number/scalar. Therefore, there exists a one-to-one correspondence between a D vectors and a complex numbers. Im{z} b z=a+bi a Re{z} b z=a bi Figure 1: A complex number z and its conjugate z in complex space. Horizontal axis contains all real numbers, vertical axis contains all imaginary numbers. The complex conjugate: z = a bi, which is obtained by reversing the sign of Im{z}. Notice that: Re{z} = z + z = a, Im{z} = z z i = b. Therefore, both Re{z} and Im{z} are linear combinations of z and z. 4

5 5. Basic computations between complex numbers Addition/subtraction: If z 1 = a 1 + b 1 i, z = a + b i (a 1, a, b 1, b R), then z 1 ± z = (a 1 + b 1 i) ± (a + b i) = (a 1 ± a ) + (b 1 ± b )i. Or, real parts plus/minus real parts, imaginary parts plus/minus imaginary parts. Multiplication by a real scalar α: αz 1 = αa 1 + αb 1 i. Multiplication between complex numbers: z 1 z = (a 1 +b 1 i)(a +b i) = a 1 a +a 1 b i+a b 1 i+b 1 b i = (a 1 a b 1 b )+(a 1 b +a b 1 )i. All rules are identical to those of multiplication between real numbers, just remember that i = 1. 5

6 Length/magnitude of a complex number z = a + bi z = z z = (a + bi)(a bi) = a + b, which is identical to the length of a D vector (a, b). Division between complex numbers: z 1 = z 1 z = (a 1 + b 1 i)(a b i) = (a 1a + b 1 b ) + (a b 1 a 1 b )i z z z z a +. b Eg 5..1 Given that z 1 = 3 + 4i, z = 1 i, calculate 1. z 1 z ;. z 1 ; 3. z 1 ; 4. z z 1. Ans: 1. z 1 z = (3 1) + (4 ( ))i = + 6i;. z 1 = i = i; 3. z 1 = z 1 z 1 = = 5 = 5; 4. z z 1 = z z 1 z 1 z 1 = (1 i)(3 4i) 5 = 5 10i 5 = i. 6

7 5.3 Complex-valued exponential and Euler s formula Euler s formula: e it = cos t + i sin t. (3) Based on this formula and that e it = cos( t)+i sin( t) = cos t i sin t: cos t = eit + e it, sin t = eit e it. (4) i Why? Here is a way to gain insight into this formula. Recall the Taylor series of e t : e t = n=0 t n n!. Suppose that this series holds when the exponent is imaginary. e it = n=0 (it) n n! = n even (it) n n! + n odd (it) n n! = (it) m (m)! + (it) m+1 (m + 1)! = i m t m (m)! + i m+1 t m+1 (m + 1)! = (i ) m t m (m)! + i(i ) m t m+1 (m + 1)! 7

8 = ( 1) m t m (m)! + i ( 1) m t m+1 (m + 1)! = cos t + i sin t. Remarks: Sine and cosine functions are actually linear combinations of exponential functions with imaginary exponents. Similarly, hyperbolic sine and cosine functions are linear combinations of exponential functions with real exponents. sinh(t) = et e t, cosh(t) = et + e t. 8

9 5.4 Polar representation of complex numbers For any complex number z = x + iy ( 0), its length and angle w.r.t. the horizontal axis are both uniquely defined. Im{z} y ρ= x + y θ z=x+yi = ρ e iθ Re{z} x Figure : A complex number z = x + iy can be expressed in the polar form z = ρe iθ, where ρ = x + y is its length and θ the angle between the vector and the horizontal axis. The fact x = ρ cos θ, y = ρ sin θ are consistent with Euler s formula e iθ = cos θ + i sin θ. One can convert a complex number from one form to the other by using the Euler s formula: z = x + iy z = ρe iθ, where x = ρ cos θ, y = ρ sin θ; ρ = x + y, θ = tan 1 y x ; where we often restrict 0 θ π or π θ π. Otherwise, the conversion from Cartesian to polar coordinates is not unique, θ can differ by an integer multiple of π. 9

10 Eg Convert the following complex numbers from one form to the other. 1. z = 3i;. z = 1; 3. z = 1 + i 3; 4. z = i; 5. z = e iπ 6 ; 6. z = 5e iπ 4 ; 7. z = 5e iπ 3. Ans: 1. z = 3i = 3e iπ ;. z = 1 = e i0 = 1, (for a positive real number, there is no change!); 3. z = 1 + i 3 = 1 + ( 3) e i tan = e iπ 3 ; 4. z = i = ( ) + ( ) e i tan 1 ( ) = e i5π 4 = e i3π 4 ; 5. z = e iπ 6 = cos( π 6 ) + i sin( π 6 ) = 3 i1 ; 6. z = 6e iπ 4 = 6 cos π 4 + i6 sin π 4 = i = 3 (1 + i); 7. z = 4e iπ 3 = ( 4) [ [ ] cos( π 3 ) + i sin( π 3 )] 1 = ( 4) i + i. = 10

11 Remark: It is important to know that the collection of all complex numbers of the form z = e iθ form a circle of radius one (unit circle) in the complex plane centered at the origin. In other words, the equation for a unit circle centered at the origin in complex plane is z = e iθ (see figure). Im{z} ρ = 1 θ z = e iθ Re{z} Figure 3: The collection of all complex numbers of the form z = e iθ form a unit circle centered at the origin in the complex plane. Remark: Rotation of a vector represented by a complex number z = ρe iθ counter-clockwise by angle φ is achieved by multiplying e iφ to it: e iφ z = e iφ ρe iθ = ρe i(θ+φ). Remark: The product between z 1 = ρ 1 e iθ 1 and z = ρ e iθ yields z 1 z = ρ 1 e iθ 1 ρ e iθ = (ρ 1 ρ )e i(θ 1+θ ) which is a vector of length ρ 1 ρ and an angle θ 1 + θ. 11

12 Eg 5.4. Find all roots of 3 1, complex and real. Ans: Therefore, 1 = e 0 = e i(0) = e (πn)i, for any integer n. 3 1 = = ( e (πn)i )1 3 = e πn 3 i, for any integer n. Often, the angle θ for a complex number expressed in form of e θi is restricted in the range 0 θ < π. If so, 3 1 has only three roots in this range: 3 1 = 1, e π 3 i, e 4π 3 i. Im{z} 3 1 = e π 3 i 3 1 = 1 Re{z} 3 4π 1 = e 3 i Figure 4: The cubic roots of number 1 in complex plane. 1

13 5.5 Polynomials of degree n must have n roots! Eg Find all roots of z + z + 10 = 0. Ans: Notice that z + z + 10 = z + z = (z + 1) + 9 = 0. There is no real root! But there are two complex-valued roots forming a pair of complex conjugates. (z +1) +9 = 0 (z +1) = 9 z +1 = ± 9 z = 1±3i. Final remarks: (a) Any polynomial of degree n can always be factored into the product of n terms in which z i, (i = 1,..., n) are the n roots. P n (z) = a n z n +a n 1 z n 1 + +a 1 z+a 0 = a n (z z n )(z z n 1 ) (z z 1 ). (b) Complex roots of a polynomial always occur as a pair of complex conjugates: z ± = a ± bi. 13