Complex Numbers : Solutions David W.H. Swenson Exercise 1. What Cartesian point is equivalent to the complex number 6i? What about −2? Since 6i = 0+6i, we identify a = 0 and b = 6 in a+bi. Therefore, (a, b) becomes (0, 6). Similarly, −2 gives us −2 + 0i, and the point (−2, 0). Exercise 2. For the complex number z = a + ib, what is |z| in terms of a and b? [Hint: think back to trigonometry.] If we look at figure 1, we see that |z|, which is denoted r in the figure, is the hypotenuse of a right triangle √with side lengths a and b. So, by the Pythagorean theorem, we have |z| = r = a2 + b2 . Exercise 3. For z = a + ib, what is arg(z) in terms of a and b? For the special case of a real number (b = 0) what is arg(z)? This is again just a little trigonometry. We know that arg(z) is the angle identified in figure 1, and we again use the fact that we have a right triangle. With b as the length of the side opposite the angle, and a the side adjacent, we have tan(arg(z)) = b/a, or arg(z) = tan−1 (b/a). There is a subtlety to the special case of a real number. The obvious answer is that arg(z) = tan−1 (0) = 0. But what about negative numbers? Since r has to be positive (since it is a distance), using arg(z) = 0 only includes the positive numbers. From looking at figure 1, we can determine that we also need to include the possibility arg(z) = π. The reason is that the function tan(θ) is π-periodic. So for any n ∈ Z, we have tan(arg(z)) = 0 ⇒ arg(z) = nπ This means that arg(z) = 0 is only the solution for n = 0. Other valid solutions include ±π, ±2π, . . .. Positive real numbers are covered by even multiples of π (including 0), and negative numbers are covered by odd multiples. Exercise 4. For a complex number z with magnitude r and argument θ, what are a and b such that z = a + ib? One more flashback of trig: defining θ ≡ arg(z) and r ≡ |z|, we have a = r cos(θ) and b = r sin(θ). Exercise 5. Show that (a + ib)(a − ib) = a2 + b2 1 Just a little algebra: (a + ib)(a − ib) = a2 − iab + iab − i2 b2 = a2 + 0 − (−1)b2 = a2 + b2 Exercise 6. Show that (r eiθ )(r e−iθ ) = r2 Remember that when you multiply exponentials, the exponent adds. So we have (r eiθ )(r e−iθ ) r2 eiθ e−iθ r2 eiθ−iθ r2 e0 r2 = = = = Exercise 7. What is the complex conjugate of a real number? For a real number, we can write z = a + 0i = a for some real number a. So the complex conjugate z ∗ = a − 0i = a, which is also equal to z. So a real number is its own complex conjugate. [Suggestion : show this using Euler’s z = r eiθ representation of complex numbers.] Exercise 8. Take a point in the complex plane. In the Cartesian picture, how does the act of taking the complex conjugate move the point? What about in the polar coordinate picture? In the Cartesian picture, we have a+ib, which becomes a−ib. This is equivalent to taking the point (a, b) and moving it to (a, −b), which is a reflection about the x-axis. So in the Cartesian picture, complex conjugation is a reflection about the x-axis. In the polar coordinate picture, we change an angle from positive to negative. So in the polar coordinate picture, complex conjugation changes the angle counter-clockwise to clockwise. You should think about these two pictures, and convince yourself that they are equivalent. Exercise 9 (Advanced). Prove Euler’s formula. [Hint : what’s the Taylor series (or MacLaurin series, actually) of ex ? So what if you replace x by iθ (remembering that i2n = (−1)n )? Now what are the series expansions for cos(θ) and sin(θ)?] We begin with the MacLaurin series of ex : ex = ∞ X xj j=0 2 j! Now we make the change of variables x → iθ: iθ e = = ∞ X (iθ)j j=0 ∞ X j=0 = = = j! (iθ)2j+1 (iθ)2j + (2j)! (2j + 1)! ∞ 2j 2j X i θ j=0 ∞ X j=0 ∞ X (2j)! = j=0 i2j+1 θ2j+1 (2j + 1)! i2j θ2j+1 −1j θ2j +i (2j)! (2j + 1)! −1j θ2j+1 θ2j + i(−1)j (2j)! (2j + 1)! −1j ∞ X θ2j θ2j+1 −1j +i (2j)! (2j + 1)! j=0 j=0 ∞ X + = cos(θ) + i sin(θ) The first step splits the sum into even and odd terms. The rest is just manipulation until the last step indentifies the series expansions found as those of sin(θ) and cos(θ). Strictly speaking, we have to prove that all these series converge over what is called “an infinite radius of convergence.” But we’ll leave that problem to the folks who have taken complex analysis. Exercise 10. Using Euler’s Formula, show that the simple rule for complex conjugation gives the same results in either real/imaginary form or magnitude/argument form. [Hint: take a complex number z = reiθ and define a and b such that reiθ = a + ib. Then take the complex conjugate.] Using the results from exercises *** and ***, for z we have a = r cos(θ) and b = sin(θ) in Cartesian (real/imaginary) form. For z ∗ = r e−iθ , a = r cos(−θ) = r cos(θ) and b = r sin(−θ) = −r sin(θ) in Cartesian form. Comparing these, we have that a for z equals a for z ∗ and b for z equals −b for z ∗ . Exercise 11. Two other formula are often grouped in with Euler’s formula. They are: 1 iθ cos(θ) = e + e−iθ 2 and 1 iθ sin(θ) = e − e−iθ 2i Prove these using Euler’s formula as given in equation ??. [Hint: sin(−x) = − sin(x) and cos(−x) = cos(x).] 3 The trick is to use Euler’s formula twice. For the positive angle, we have eiθ = cos(θ) + i sin(θ) and for the negative angle, we have e−iθ = = cos(−θ) + i sin(−θ) cos(θ) − i sin(θ) where the second step comes from the parity (even/odd-ness) of the sin and cos functions, which was given in the hint. Now all we have to do is either add or subtract the functions. If we add them, we find eiθ + e−iθ = (cos(θ) + i sin(θ)) + (cos(θ) − i sin(θ)) = 2 cos(θ) From that, we get 12 eiθ + e−iθ = cos(θ). On the other hand, if we subtract them, we find eiθ − e−iθ (cos(θ) + i sin(θ)) − (cos(θ) − i sin(θ)) 2i sin(θ) 1 And from there we easily obtain 2i eiθ − e−iθ = sin(θ). = = Exercise 12 (Advanced). There’s a famous formula in mathematics which combines several of the most important mathematical constants: e, π, i, and 1. Construct a formula which is equal to zero, using each of those constants once in your expression. [Hint : remember that θ in eiθ is in radians.] We’ll go straight to the answer: eiπ + 1 = 0. A friend bought me a pin with this formula on it. Convince yourself that it is true. Exercise 13. What is the square root of i? Following the methodology outlined in the text, we first convert i to Euler’s notation. It has modulus 1 and argument π/2. So p √ i = eiπ/2 √ Now we use the fact that z = z 1/2 and we have 1/2 √ 1 i = eiπ/2 = ±eiπ/2 2 = ±eiπ/4 Since ±1 = ei(π/2∓π/2) , we have √ i = = = = ei(π/2∓π/2) eiπ/4 ei(π/2∓π/2+π/4) ei(3π/4∓π/2) eiπ/4 or e5π/4 4 Converting these back to real part/imaginary part notation: π π i 1 + i sin =√ +√ eiπ/4 = cos 4 4 2 2 and 5π 1 i 5π e = cos + i sin = −√ − √ 4 4 2 2 This exercise is part of an interesting subject in mathematics called the nth roots of unity. M&S give more detail in their exercise ***. 5iπ/4 Exercise 14. Prove de Moivre’s formula, n (cos(θ) + i sin(θ)) = cos(nθ) + i sin(nθ) where θ ∈ R and n ∈ N. [Hint : (eb )c = ebc ] Once we have Euler’s formula, this is pretty straightforward. Thanks to Euler, we have n n (cos(θ) + i sin(θ)) = eiθ = e(iθ)n = ei(θn) = cos(nθ) + i sin(nθ) This also reminds us of another important rule when dealing with exponentials, which was given in the hint. Exercise 15 (Advanced). The technique described above can be used to find many trigonometric identities. By first taking the trig function, then using the formulae given by equations ?? and ??, doing some math with the result, then converting them back to trigonemetric forms, you can rather easily obtain many results from trigonometry. As an example, try sin2 (θ) + cos2 (θ) = 1 (To the real show-offs: try R 1 dx sin2 (ax) cos2 (ax) = − 32a sin(4ax) + x8 ) As the exercise suggests, we first replace the sin and cos functions, then do some arithmetic: iθ 2 iθ 2 e − e−iθ e + e−iθ sin2 (θ) + cos2 (θ) = + 2i 2 2 1 1 2 = eiθ − e−iθ + eiθ + e−iθ −4 4 2 1 iθ −iθ 2 = − e −e + eiθ + e−iθ 4 1 = − e2iθ − 2 + e−2iθ + e2iθ + 2 + e−2iθ 4 1 −e2iθ + 2 − e−2iθ + e2iθ + 2 + e−2iθ = 4 1 = 4=1 4 5 Now, for the show-off version. I’ll do the math a little less explicitly, but you should be able to at least follow the gist: Z 2 2 dx sin (ax) cos (ax) 2 1 iax 2 1 iax −iax −iax e −e e +e dx 2i 2 Z 1 2iax dx − e − 2 + e−2iax e2iax + 2 + e−2iax 16 Z 1 4iax dx − e − 2 + e−4iax 16 Z Z Z 1 4iax −4iax − dx e − dx 2 + dx e 16 1 1 4iax 1 −4iax − e − 2x + e 16 4ia −4ia x 1 1 − e4iax − e−4iax + 16 4ia 8 x 1 sin(4ax) + − 32a 8 Z = = = = = = = 6