Section 2.2

Question 9: (a) (z-5i)^2 is a polynomial, hence continuous everywhere, and thus on e can simply substitute in the limit to obtain ((2+3i)^2 - 5i)^2 = -8i.

(c) This is an indeterminate form; one can either use L'hopital's rule (which is valid for complex functions, although we will not prove this until much later), or factorize the numerator as z^2 + 9 = (z-3i)(z+3i) and simplify the limit to z+3i. Now the function is continuous and we can simply substitute to obtain 6i.

(e) The function in the limit can be simplified to 2z_0 + /\z, which (being a polynomial in /\z) converges to 2z_0 as /\z tends to zero. Alternatively, one can note that this is precisely the derivative of z^2 at z_0, which is equal to 2z_0 since the differentiation of polynomials proceeds in complex analysis in exactly the same manner as in real analysis.

Question 15: The limit along the real approach y=0 is 2i, and the limit along the imaginary approach x=0 is 2i+1, which are not equal, so the full limit does not exist.

Section 2.3

Question 4: (a) The expression in the limit of Definition 4 simplifies to /\x / /\ z, where /\z = /\x + i /\y is the Cartesian form of /\z. Taking the limits along the real and imaginary approaches shows that this limit does not exis t.

(b) is proven in the same way as (a).

(c) is trickier, and one needs to divide into two cases, z_0 = 0 and z_0 != 0. Wh en z_0 = 0 the argument is similar to (a) and (b). When z_0 is nonzero one needs a t rick to simplify the expression, namely to multiply the numerator and denominator by |z_0 - /\z| + |z_0|. One can use the identity |z|^2 = zz* to simplify the numerat or, while the |z_0 - /\z| + |z_0| term in the denominator converges to 2|z_0| and can be taken out of the limit. The remaining terms in the expression will eventually simplify to 2z* + /\z + /\z* / /\z, which does not converge because the first two terms converge and the last term diverges. (The sum of a convergent limit and a divergent limit is always divergent).

Section 2.4

Question 2: Write h = u+iv. Since the partial derivatives of u and v are polynomials, they are continuous, and so differentiability is equivalent to the Cauchy-Riemann equations being satisfied. In this case these equations become

 3x^2 + 3y^2 - 3 = 3y^2 + 3x^2 - 3

 6xy = -6xy.

Question 5: The partial derivatives, being products of polynomials, exponential functions, sines and cosines, are continuous, and so it suffices to verify that the Cauchy-Riemann equations hold. Once one does this, one can use the identity df/dz = df/dx = 1/i df/dy to find the derivative. Alternatively, one can observe that f(z) can be written much more compactly as exp(z^2), so f is evidently differentiable everywhere with a derivative of 2z exp(z^2).

Question 11: Write f = u+iv. Since f is analytic in D, f is differentiable in D and so therefore f obeys the Cauchy-Riemann equations du/dx = dv/dy, du/dy = -dv/dx. Similarly f* = u-iv must also satisfy the Cauchy-Riemann equations, so du/dx = -dv/dy and du/dy = dv/dx. Combining these equations together and simplifying we see that both partial derivatives of u and v are zero, which (since D is a domain) implies that u and v are both constant, so that f is constant.

Section 2.5

Question 1: (a) f = u+iv where u(x+iy) = x^2 - y^2 + 2x + 1 and v(x+iy) = 2xy + 2y; du^2/dx^2 + du^2/dy^2 = 2 - 2 = 0 and dv^2/dx^2 + dv^2/dy^2 = 0 + 0 = 0.

(c) h = u+iv where u(x+iy) = exp(x) cos y and v(x+iy) = exp(x) sin y; du^2/dx^2 + du^2/dy^2 = exp(x) cos y - exp(x) cos y = 0 and dv^2/dx^2 + dv^2/dy^2 = exp(x) sin y - exp(x) sin y = 0/

Question 3: (c) u is the imaginary part of exp(z), so it is the real part of -iexp(z). Since -i exp(z) is evidently analytic, u is automatically harmonic, and the imaginary part of -i exp(z), -exp(x) cos(y), is a conjugate of u.

(e) u is the real part of Log(z), which is analytic when Re z > 0, so u is automatically harmonic, and the imaginary part of Log(z), which is arctan(y/x) when Re z > 0, is a conjugate of u. (You can also get this result by directly solving the Cauchy-Riemann equations, but you have to be careful of the fact that y could be zero, so one must look out for any expressions with y in the denominator.)

Section 7.3

Question 12: There are several ways to do this. One approach is as follows:

• Apply an inversion w_1 = 1/z. This maps the half-plane to the exterior of the circle with center 1/2 + 1/2 i and radius 1/sqrt(2).

• Move the center of the circle to the origin by applying the translation w_2 = w_1 - (1/2 + 1/2 i).

• Adjust the radius of the circle to 1 by applying the dilation w_3 = sqrt(2) w_1.

• Move the exterior of the unit circle to the interior of the unit circle by applying an inversion w_4 = 1/w_3.

w = z / (sqrt(2) - exp(pi i/4) z);