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NOTES on COMPLEX VARIABLES 1. the Complex Exponential

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NOTES on COMPLEX VARIABLES 1. the Complex Exponential NOTES ON COMPLEX VARIABLES 1. The complex Exponential Function ez, where z = x + iy The complex exponential function is defined by extending the Taylor series of ex from real values of x to complex values: ez =1+ z + z2/2+ z3/3! + ... (1) The partial sums are well defined since products and sums of complex numbers are well defined. With this definition, ez agrees with ex when z is real. Furthermore, one can show that ez satisfies z1 z1+z2 z1 z2 z1−z2 e z n nz d z z the properties of exponentials: e = e e , e = ez2 , (e ) = e , dz [e ]= e . We showed in class that the definition (1) implies Euler’s formula: if θ is a real number, then eiθ = cos θ + i sin θ (2) which is a point on the unit circle that subtends an angle θ with the real axis. Every complex number z in the complex plane is obtained by taking a point on the unit circle times its magnitude r. Thus, Euler’s formula enables us to write any complex number z = x + iy in the form z = reiθ = r cos(θ)+ ir sin(θ) (3) where r is the modulus and θ is the argument. We’ll refer to this form as the polar representation of z in terms of the coordinates r, θ. The relation between the polar representation z = reiθ and the Cartesian representation z = x + iy can be seen from (3) to be given by x = r cos θ, y = r sin θ and r = x2 + y2,θ = atan(y/x) . p By writing complex numbers in polar form it is easy to take powers and roots of complex numbers, as shown in the following examples. Example 1: Plot the complex number 1+ i in the complex plane and find its polar representation. Solution: By plotting the point we see that θ = π/4 and r = √2. Thus 1 + i = √2eiπ/4. Example 2: Find the real and imaginary parts of z = (1+ i)10. Solution: We use the polar representation 1 + i = √2eiπ/4 and the properties of the exponential function to find that 10 π 10 5 5π 5 π 5 z = (1+ i) =(√2ei 4 ) = 2 ei 2 = 2 ei 2 = 2 i so Re(z)=0, Im(x) = 25. That is, z = (1+ i)10 = 32i. Notice that taking a complex number z = reiθ to a power zn = rneinθ 1 simply takes the power of the modulus and multiplies the argument by n. For example squaring a number squares the modulus and doubles the argument. Exercise: Plot all integer powers of the complex number on the unit circle (1+ i)/√2. Euler’s formula also enables us to identify the real and imaginary parts of ez, where z = x + iy, as follows ez = ex+iy = exeiy = ex(cos y + i sin y)= ex cos y + iex sin y . (4) so Re(z)= ex cos y and Im(z)= ex sin y. Example 3: Find the real and imaginary parts of z = e1+i . Solution: z = e1+i = e1ei = e(cos1 + i sin1), so Re(z)= e cos1 and Im(z)= e sin1. Note that the argument of a complex number is not uniquely defined since you can add any multiple 2nπ of 2π, where n is any positive or negative integer, to θ and get the same points. That is, if z = reiθ then also z = rei(θ+2nπ) For example, the number 1 = e0i = e(0+2nπ)i = e2nπi. We’ll use this to find all complex roots of equations. In particular, all roots of unity. Example 4: Find all complex roots of z3 = 1. Note: The fundamental theorem of algebra states that a polynomial of degree n has precisely n complex roots, possibly repeated. This implies that z3 1 = 0 has precisely 3 complex roots. − Solution: By taking the modulus on both sides we find that z3 = z 3 =1 so z =1 so z must be a point on the unit circle, | | | | | | z = eiθ By plugging this into the equation and writing 1 = e2nπi we get that ei3θ = e2nπi Now we can equate powers to get for n =0: θ = 0 for n =1: θ = 2π/3 for n =2: θ = 4π/3 Any other values of n only give values of θ that agree with one of the above to within a multiple of 2π. So the 3 distinct cubic roots of unity are 2π 4π z = 1,e 3 i,e 3 i . Alternatively, one can think of this problem as follows: we want to find a number z such that the 3 i2π/3 argument of z is 2π. Since cubing it triples the argument of z we deduce that z0 = e is a 3 2 3 j solution. Since z0 = 1 it follows that any power z0 , z0 , ... , z0, ... , also solve the equation, since j 3 3 j j they satisfy (z0) = (z0 ) = 1 = 1. You can check that only 3 of those powers are distinct. z0 is called the principal root of unity. 2 2. Deriving Trig identities using Euler’s Formula Euler’s formula and the properties of exponentials make it is easy to derive several of the trigono- metric identities we have been using. Example 10: Use exponentials to derive formulas for sin2x and cos 2x. Solution: Start with the identity e2ix =(eix)2. Using Euler’s formula we rewrite this identity as cos 2x + i sin2x = (cos x + i sin x)2 = cos2 x sin2 x + 2i sin x cos x − By equating real and imaginary parts on the left and right hand side of this equation we get cos 2x = cos2 x sin2 x , sin2x = 2sin x cos x − 3. Integration using Complex Variables Complex variables can also be used to determine integrals that are tedious to evaluate using conven- tional methods. For example, consider integrals of the form eat cos bt dt. Recall how to integrate using integration by parts. Here it is shown that the use of coRmplex variables can sometimes reduce the complecity of the integration of such functions. You will need to use Euler’s equation and the definition of equality for complex numbers – namely that if z1 = z2 then Re(z1) = Re(z2) and Im(z1)= Im(z2). We first define what the integral of a complex function means. Consider a complex valued function w(t)= u(t)+ iv(t), where u and v are real-valued function and t is a real variable. (That is, u and v are the real and imaginary parts of w.) Then the integral of w is defined by b b b w(t)dt = u(t)dt + i v(t)dt (10) Za Za Za 1 2 Example 11: Evaluate the integral 0 1 + 2it dt Solution: 1 2 1 R 1 2 1 3 1 0 1 + 2it dt = 0 1 dt + 2i 0 t dt =[t]0 + 2i t /3 0 =1+2i/3 R R R It follows that the standard rules of integration can be used by treating i as a constant. π/3 2it Example 12: Evaluate the integral 0 e dt π/3 Solution: π/3 e2itdt = 1 e2it =R 1 e2iπ/3 e0 = −i 3/2+ i√3/2 = √3/4 + 3i/4 0 2i 0 2i − 2 − We nowR illustrate that some of the integrals we computed earlier this semester are evaluated quite easily if we use complex variables instead. Example 13: Evaluate the integral e(a+ib)tdt. What do you learn by equating real and imaginary parts on the left and rightR hand sides of the resulting equation? 1 Solution: e(a+ib)tdt = e(a+ib)t + C . Equating real and imaginary parts gives Z a + ib at at e e cos bt dt = (a cos bt + b sin bt)+ C1 , Z a2 + b2 at at e e sin bt dt = (a sin bt b cos bt)+ C2 , Z a2 + b2 − 3 Notice that to compute these two integrals without complex variables, we would have to use integration by parts twice. These integrals occur frequently in mechanical systems. The functions eat cos bt, eat sin bt represent oscillating functions whose magnitude grows or decays exponentially. Problems: Part I Powers of complex numbers: 1. Write the following numbers in the form reiθ and in the form a+ib: (a) (1+i)20, (b) (1 √3i)5, (c) (2√3 + 2i)5, (d) (1 i)8. − − 2. Find all solutions to the given equation. Sketch the roots in the complex plane. (a) z8 = 1 (the eigth roots of unity) (b) z5 = 1 (the fifth roots of unity) (c) z5 = 32 (the fifth roots of 32) Problems: Part II Exponentials and Logarithms: πi 2±3πi 2 2+ e z+πi z 1. Show that: (a) e = e (b) e 4 = (1 + i) (c) e = e − 2 − p 2. Find the real and imaginary parts of e(2−3i)t (t is real) Trigonometric Identities: 3. Derive the formulas for sin(a + b) and cos(a + b) using complex variables. (Hint: look at the real and imaginary parts of the equation ei(a+b) = eiaeib.) Integration: 4. Evaluate the following integrals 1 2 π/4 it (a) 0 (1 + it ) dt (b) 0 e dt 5. FindR the indefinite integralR e−st cos(at) dt, where s,a are positive constants (a) using integration by partsR twice (b) by integrating e−steiat dt, and then equating real and imaginary parts. 6. The Laplace transformR of a function f(t) is defined to be ∞ F (s)= [f](t)= e−stf(t) dt L Z0 (a) Find the Laplace transform of f(t) = cos(at).
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