Section 1.1

Question 15: The simplest way to do this is to use polar co-ordinates, but I shall use the exponent laws. (The laws of exponents are part of the laws of algebra, and as such are just as valid for the complex numbers as they are for the reals or rational numbers - as long as all t he exponents are integers. If the exponents are fractional, irrational, or complex it gets a little trickier, as we shall see later on in the course).

Since i^4 = 1, we see that i^{4k} = (i^4)^k = 1^k = 1, i^{4k+1} = i^{4k} i = 1 i = i, etc.

Question 20: All the normal rules of algebra apply:

(a) 2iz = 4, so z = 4/(2i) = -2i

(b) z = (1-5i)(1-z), so (2-5i)z = (1-5i), so z = (1-5i)/(2-5i) = (-23 - 15i)/29 = -23/29 - 15i/29.

(c) z(8z + (2-i)) = 0, so either z = 0 or z = -(2-i)/8 = -1/4 + i/8.

(d) z^2 = -16, so z = +4i or -4i.

For the purposes of this course, "solving for z" means expressing z in one of the two standard forms (Cartesian or polar). Unless the problem asks you to simplify, however, then you may get away with an unsimplified form such as (1-5i)/(2-5i). Just don't count on it.

Question 23: Write z=x+iy in Cartesian form. If Re(z)>0, then x>0. Now 1/z in Cartesian form is x/(x^2+y^2) - iy/(x^2 + y^2), so Re(1/z) = x/(x^2+y^2), which is obviously positive since x>0 by assumption and x^2+y^2 is positive (it can't be zero since x is non-zero).

Section 1.2

Question 7: Answers in textbook. For (c), divide both sides of the equation by 2 and interpret the resulting equation geometrically. For (d), either use geometry or write z in Cartesian form and simplify. For (h), interpret the real part of z geometrically.

Section 1.4

Question 4: (a) The most obvious polar form is exp(2 pi i/9)^3 = exp(2 pi i/3).

(b) 2+2i = sqrt(8) exp(pi i / 4) and -sqrt(3) + 1 = 2 exp(2 pi i / 3), so the quotient is sqrt(2) exp(5 pi i / 8). Other phases are possible.

(c) 2i = 2 exp(pi i/2), and 3exp(4+i) = 3e^4 exp(i), so the quotient is 2/(3e^4) exp(i (pi/2 - 1)).

Question 17: Geometrically, for 0 <= t <= 2pi, e^{it} is the point on the unit circle whose counterclockwise angle with the positive real axis is t. As t goes from 0 to 2pi, this point therefore traverses the unit circle once in the counterclockwise direction. The answers to (a)-(d) are in the textbook, and can be seen by similar arguments to the above.

Section 1.5

Question 5: (a) -16 = 16 exp(pi i + 2kpi i), so the fourth roots of -16 are 2 exp(i(pi + 2k pi)/4) for k = 0,1,2,3 (all the other values of k are redundant).

(d) 1-sqrt(3) i = 2 exp(-pi i /3 + 2k pi i), so the cube roots are 2^{1/3} exp(i(-pi/3 + 2k pi)/3) for k = 0,1,2 (all the other values of k are redundant).

Section 1.6

Questions 2-5: The set (b) describes an infinite sector which does not contain i ts boundary, and is therefore open, unbounded, and connected, hence a domain.

The set (c) describes a disk with its center and boundary removed, and is theref ore open, bounded, and connected (although not simply connected), hence a domain.

The set (d) describes an infinite strip which only contains one of its boundary lines, and is therefore neither open nor closed, unbounded, connected, but not a domain .

Section 2.1

Question 1(f): exp(z) = exp(x) cos(y) + i exp(x) sin(y), and exp(-z) = exp(-x) c os(y) - i exp(-x) sin(y), so G(z) = u(x,y) + i v(x,y), where u(x,y) = (exp(x) + exp(-x )) cos(y) = 2 cosh(x) cos(y) and v(x,y) = (exp(x) - exp(-x))sin(y) = 2 sinh(x) sinh(y).