Let [`z] = z<sup>3</sup>. Then |z| = |z|<sup>3</sup> so |z| = 0 or |z| = 1. One solution is z = 0. All the other solutions have |z| = 1 and, for them, z<sup>4</sup> = [`z]z = |z|² = 1. This leads to four other solutions, z = 1,i,-1,-i.

Alternatively, if we assume that the given complex number is of the form x + iy, where x and y are real, we are led to the equations x(x² - 3y² -1) = 0,  y(3x² -y² +1) = 0. This leads to four possibilities: (i) x = 0, y = 0; (ii) x = 0,  3x² -y² +1 = 0; (iii)x² - 3y² -1 = 0,  y = 0; (iv) x² - 3y² -1 = 0,  3x² -y² +1 = 0. The first of these leads to z = 0, the second to z = ±i, and the third to z = ±1. Adding the two equations in the fourth leads to x² = y² and putting this back in either of the equation leads to an impossible equation 2y² + 1 = 0.

It is clear that |z_n - 0| = ||z_n| - 0|, so |z_n - 0| is small if and only if ||z_n| - 0| is small. Hence lim_{n ®¥} z_n = 0 is equivalent to lim_{n ® ¥} |z_n| = 0. On the other hand, |z_n - 1| ¹ ||z_n| - 1|, and in fact ||z_n| - 1| can be small without |z_n - 1| being small. For example, we could have z_n = -1, for each n. Then lim_{n® ¥} |z_n| = 1 but lim_{n® ¥} z_n = -1.

The function fails to be differentiable for z ¹ 0, because the Cauchy-Riemann equations do not hold. On the other hand,

• Evaluate ò_C Re  z  dz, where C is the line segment joining 0 and 1+i.

  C is described by the parametric equation z(t) = (1+i)t,  0 £ t £ 1, and on this curve, Re z = t,  z¢(t) = 1+i. This the given integral is equal to

  ó õ 1 0  t(1+i)  dt = (1+i)/2.

• Evaluate ò_C |z|  dz where C is the line segment joining -1 to 1. C is described by the parametric equation z(t) = t,  -1 £ t £ 1, and on this curve, |z| = |t|,  z¢(t) = 1. This the given integral is equal to

  ó õ 1 -1  |t|  dt = 1.

Cauchy's Integral Theorem: Let f(z) be analytic in a simply connected domain G. Then ò_C f(z) dz = 0 for every piecewise smooth closed curve C contained in G. (The phrase simply connected is necessary; for example, 1/z is analytic in the domain {z|z ¹ 0} but its integral around the circle |z| = 1 is not 0.)

Cauchy's Integral Formula:

Let f(z) be analytic be a domain G containing a piecewise smooth closed curve C and its interior. Then

Cauchy's Integral Theorem shows that under appropriate conditions an integral of a function is independent of path. This makes possible the idea of an indefinite integral and leads to the possibility of deforming a contour and Cauchy's Integral Formula which shows, among other things that an analytic function is determined by its values in a relatively small set of points and that an analytic function is infinitely differentiable.

Let s_n(x) = xⁿ,  n = 1,2.... Then s_n(x) ® f(x),  0 £ x < 1, where

(a) å_{n = 1}^¥ zⁿ. Here a_n = zⁿ and

(b) å_{n = 0}^¥ zⁿ/n!. Here a_n = zⁿ/n! and