Midterm 1 Solutions

This is false. The principal argument always has to be in the range from -pi to pi, and the sum of two numbers in this range is not necessarily also in that range.

Question 1(b)

This is false. In fact, if we write z=x+iy and w=u+iv, then (x+iy)(u+iv) = (xu-yv) + i(xv+yu), so we see that the real part Re(zw) equals xu-yv = Re(z)Re(w) - Im(z)Im(w). This will not in general be equal to Re(z)Re(w), unless the product Im(z)Im(w) happens to be zero.

Question 1(c)

This is true:

                  __               __
             _     _      _   _    _    _
       _    zw  + zw     zw + zw   zw + zw      _
   Re(zw) = ---------  = ------- = ------- = Re(zw).
                2           2         2

Question 1(d)

This is true. By the triangle inequality,

    _    _      _      _     _         _              
   |z - zw| <= |z| + |zw| = |z| + |z| |w| = |z| + |z| |w| = |z|(1 + |w|).

Question 1(e)

This is false. For example, if z=1 and w=i,

       _                         _
   Im(zw) = Im(-i) = -1  but  Im(zw) = Im(i) = 1.

Question 2

The principal argument of -sqrt(3)+i is 5pi/6 (draw a picture to convince yourself of this) and its modulus is 2. Therefore, -sqrt(3) + i = 2 e^(5pi/6).

The principal argument of (i-1)/7 is the same as that of -1+i, which is 3pi/4 (draw a picture to convince yourself of this). The modulus is |(i-1)/7| = sqrt(2)/7. Therefore, (i-1)/7 = (sqrt(2))/7 e^(3pi/4).

The principal argument of -4 is pi. Its modulus is 4. Therefore, -4 = 4 e^(ipi).

You could also have chosen values of the arguments other than the principal argument, as long as in part (b) you correctly identified which was the principal argument.

Question 3(a)

(1-i)^(20) = [sqrt(2)e^(-ipi/4)]^(20) = 2^(10)e^(-20ipi/4) = 1024 e^(-5pi i) = 1024 (cos(-5pi) + i sin(-5pi)) = 1024(-1 + 0i) = -1024.

Question 3(b)

This factors as z(z^4 + 16) = 0, so the solutions are z=0 together with the fourth roots (-16)^(1/4). Since -16 = 16 e^(ipi), these roots are 2 e^(ipi/4), 2 e^(ipi/4 + 2 pi i/4), 2 e^(ipi/4 + 4 pi i/4), and 2 e^(ipi/4 + 6 pi i/4).

You can either calculate each of these separately or remember that you can rewrite these as 2 e^(ipi/4) (e^(2pi i/4))^n (n = 0, 1, 2, or 3). Then calculate 2 e^(i pi/4) = 2( cos(pi/4) + i sin(pi/4)) = sqrt(2) + isqrt(2) and e^(2pi i/4) = e^(pi i/2) = i, so these roots are

sqrt(2) + isqrt(2), i(sqrt(2) + isqrt(2)), i^2(sqrt(2) + isqrt(2)), and i^3(sqrt(2) + isqrt(2))

+/- sqrt(2) +/- isqrt(2).

Question 4

                             _____        _____
      _ 2     _   2        _    _     _    _
   |z+w|  -  |z-w|   =  (z+w)(z+w) - (z-w)(z-w)
   
                           _  _       _      _
                     =  (z+w)(z+w) - (z-w)(z-w)
   
                        _        __   _    _    __         _
                     = zz + zw + wz + ww - zz + zw + wz - ww
   
                                __ 
                     = 2 zw + 2 zw
   
                              __
                         zw + zw
                     = 4 -------
                            2
   
   
                     = 4 Re(zw).

Question 5

When z is real, |z| = +/- z, so |z|^2 = z^2, and hence (z^2)/(|z|^2) = 1.

When z = iy is pure imaginary, (z^2)/(|z|^2) = (-y^2)/(y^2) = -1.

Therefore, (z^2)/(|z|^2) stays at 1 as z -> infinity along the real axis, but stays at -1 as z -> infinity along the imaginary axis. Therefore, it cannot have a well-defined limit as z -> infinity .

This proves that the limit in question does not exist.

Question 6

Write f in terms of its real and imaginary parts: f(x,y) = u(x,y) + iv(x,y). Since f is entire, the Cauchy-Riemann equations tell us that u_x = v_y and u_y = -v_x. Since f is real-valued, v(x,y)=0 and so u_x = v_y = 0 and u_y = -v_x = 0. Since the partial derivatives u_x and u_y are equal to zero everywhere, u(x,y) must be a constant function, and therefore so must f since f(x,y) = u(x,y) + 0i.

Question 7(a)

Write f(x+iy) = (x-y)^2 + 2x^2 + ( (x-y)^2 + 2y^2) i = u(x,y) + i v(x,y) where u(x,y) = (x-y)^2 + 2x^2 and v(x,y) = (x-y)^2 + 2y^2.

The partial derivatives of these are u_x = 2(x-y) + 4x = 6x - 2y, u_y = -2(x-y) = 2y - 2x, v_x = 2(x-y) = 2x - 2y, and v_y = -2(x-y) + 4y = 6y - 2x.

Because these partial derivatives are continuous everywhere, f is differentiable precisely at those points where the Cauchy-Riemann equations are satisfied. These equations are u_x=v_y and u_y = -v_x. The first equation is 6x - 2y = 6y - 2x, which is equivalent to x=y. The second equation is 2y - 2x = - (2x - 2y), which is satisfied everywhere.

Therefore, the set of points were f is differentiable is the set of points x+iy where y=x. (This is a line through the origin with slope 1).

Question 7(b)

In order for f to be analytic at a point z, it must be differentiable on some disk centred at z. But we know f is differentiable only on a single line. There is no disk (of non-zero radius) that can be completely contained within that line. Therefore, there is no disk on which f is differentiable, and hence there is no point at which f is analytic.