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Introductory Problems
1. Solve z2 + z = 1 + i
Answer
1. z = i and z = -1 – i
Solution
1. Apply the quadratic formula to z2 + z + 1 – i = 0, obtaining z= (-1 ± (-3 + 4i)1/2)/2 To compute the square root, we find the polar representation of -3 + 4i. The modulus is ((-3)2 + 42)1/2= 5. The argument is θ = arctan(-4/3). Then (-3 + 4i)2 has modulus √5 and argument θ/2. This gives: z= (-1 ± √5(cos(θ/2) + i sin(θ/2)/2 = (-1 ± √5(cos(θ/2)) + i √5 sin(θ/2)/2. But we can do better. Using the half-angle formulas: cos(θ/2) = ((1 + cos(θ))/2)1/2 and sin(θ/2) = ((1 – cos(θ))/2)1/2 we need to find cos(θ) when tan(θ) = -3/4. Recalling we started with the polar representation of -3 + 4i, we see cos(θ)= -3/5.
2. Find Cartesian expressions for (a) ei and (b) e1+i.
Answer
1 The answers are the following: (a) ei = cos(1) + i sin(1) (b) e1+i = e⋅cos(1) + i e⋅sin(1)
Solution
1. The solutions are as follows: (a) Apply Euler’s formula with z = 1: ei = ei1 = cos(1) + i sin(1). (b) e1 + i = e1ei = e(cos(1) + i sin(1)).
3. Find Cartesian expressions for (a) cos(i) and (b) cos(1+i).
Answer
1. Th answers are as follows: (a) (e1 + e-1)/2 (b) cos(1 + i) = cos(1)(e1 + e-1)/2 + i sin(1)(e1 – e-1)/2
Solution
1. The solutions are as follows: (a) From the power series for cos(z) we see cos(i) = 1 – i2/2! + i4/4! – i6/6! + i8/8! – … = 1 + 1/2! + 1/4! + 1/6! + 1/8! + … (e1 + e-1)2 (b) Use the angle sum formula for cosine to find cos(1 + i) = cos(1)cos(i) + sin(1)sin(i). We know all the parts of this except sin(i), which we find from its power series sin(i) = i – i3/3! + i5/5! – i7/7! + … = i(1 + 1/3! + 1/5! + 1/6! + … ) = i(e1 – e-1)/2 Combining these, we see cos(1 + i) = cos(1)(e1 + e-1)/2 + i sin(1)(e1 – e-1)/2.
4. Describe the curves: (a) ex+ iπ/2 and ex + iπ for all x (b) e0 + iy and e1 + iy for all y
Answer
1. The answers are the following: (a) ex + iπ/2 is the positive imaginary axis; ex + iπ is the negative real axis. (b) e0+ iy is the circle of radius 1 centered at the origin; e01+ iy is the circle of radius e centered at the origin
Solutions
1. We use ex +yi= ex(cos(y) + i sin(y)). (a) ex + iπ/2 = ex(cos(π/2) + i sin(π/2)) = exi, the positive imaginary axis. ex + iπ= ex(cos(π) + i sin(π))= ex(-1), the negative real axis. (b) e0 + iy = e0(cos(y) + i sin(y)) = cos(y) + i sin(y), the circle of radius 1 centered at the origin. e1 + iy = e1(cos(y) + i sin(y)) = the circle of radius e centered at the origin.
5. Find the period of the complex cosine function.
Answer
1. 2π
Solution
1. By the angle sum formula sin(a + b) = sin(a)cos(b) + cos(a)sin(b) we see sin(z + π/2) = cos(z). From this we see that sin(z) and cos(z) have the same period, call it α. Then by Euler’s formula, ei(z + α)= cos(z + α) + isin(z + α) = cos(z) + i sin(z) = eiz From this we see: ez + iα= ei(-iz + α)= ei(-iz)= ez That is, ez is periodic with period iα, so α= 2 π.