The differentiation formulas for sums, differences, products, quotients, integral powers, and composite functions are the same as for real-variable calculus, so these derivatives are the same as they would be for real-variable calculus.
(a) f'(z) = 6z - 2
(b) f'(z) = 3(1-4z^2)^2(-8z)
(c)
(2z+1)(1) - (z-1)(2) 3
f'(z) = --------------------- = -------
2 2
(2z+1) (2z+1)
(d)
2 2 3 2 4 2 3 2
z 4(1+z ) (2z) - (1+z ) 2z 2(1+z ) (3z - 1)
f'(z) = ---------------------------- = -----------------
4 3
z z
Page 47 #2
(a) This follows immediately from the theorems on derivatives of integer powers, sums, and products.
(b) P(0) = a_0 + a_1(0) + a_2(0^2) + ... + a_n(0^n) = a_0, P'(0) = a_1 + 2 a_2(0) + 3a_2(0^2) + ... + n a_n 0^n = a_1, and so on. (A rigorous proof would require induction, but I won't write out all the details.)
Page 48 #7 (6th edition: #9)
For ease of notation (especially for those without graphical web browsers) I'm going to use h in place of the Delta z in the book.
Recall that f'(0) exists if and only if
f(0+h) - f(0)
lim -------------
h->0 h
exists. From the
definition of
f we see that f(0)=0 and
_ 2 f(0+h)-f(0) _ 2 2
f(0+h) = f(h) = (h) / h, so ------------ = (h) / h
h
and it was proven in class that the limit of this quantity as h
goes to zero does not exist.
If you forget the discussion in class, here it is again:
as h goes to zero
along the real or imaginary axes,
_ _ 2 2 h = h or -h, so (h) / h = 1;but as h goes to zero along the line y=x (so h is a complex number of the form x+ix),
_ _ 2 2 2 2 2 2 h = x-ix, so (h) / h = (x-ix) / (x+ix) = (1-i) / (1+i) = -1which is different from 1. Since the quantity approaches different values as h -> 0 in different directions, the limit does not exist, and hence f'(0) does not exist.
Page 48 #8(b,c)
These are done in the same manner as part (a) (which was done in class).
(b) The question of whether or not f'(z) exists is the question of whether or not the limit (as h-> 0) of (Re(z+h)-Re(z))/h = (Re(z) + Re(h) - Re(z))/h = (Re(h))/h exists.
As h-> 0 along the real axis, Re(h)=h so (Re(h))/h = 1. As h-> 0 along the imaginary axis, Re(h)=0 so (Re(h))/h=0.
Therefore, (Re(h))/h approaches different numbers as h approaches zero from different directions, so the limit as h -> 0 does not exist, so f'(z) does not exist. The above argument did not depend on the value of z, proving that f'(z) does not exist for any z.
(c) The question of whether or not f'(z) exists is the question of whether or not the limit (as h-> 0) of (Im(z+h)-Im(z))/h = (Im(z) + Im(h) - Im(z))/h = (Im(h))/h exists.
As h-> 0 along the real axis, Im(h)=0 so (Im(h))/h = 0. As h-> 0 along the imaginary axis, Im(h)=h/i so (Im(h))/h=1/i.
Therefore, (Im(h))/h approaches different numbers as h approaches zero from different directions, so the limit as h -> 0 does not exist, so f'(z) does not exist. The above argument did not depend on the value of z, proving that f'(z) does not exist for any z.
Page 54 #1
(b) f(x+iy) = x + iy - (x - iy) = 0 + 2iy so u(x,y)=0 and v(x,y)=2y. This means u_x = u_y = v_x = 0 while v_y = 2 != 0. Therefore, the Cauchy-Riemann equation u_x = v_y is not satisfied for any value of z, so f'(z) does not exist at any point.
(d) f(x+iy) = e^x (cos y - i sin y) so u(x,y) = e^x cos y and v(x,y) = - e^x sin y. This means u_x = e^x cos y while v_y = - e^x cos y, so the Cauchy-Riemann equation u_x = v_y is only satisfied when e^x cos y = 0, which only happens when cos y = 0.
Also, u_y = - e^x sin y while -v_x = e^x sin y, so the Cauchy-Riemann equation u_y = -v_x is only satisfied when e^x sin y = 0, which only happens when sin y = 0.
Since sin y and cos y are never both zero, there is no value of z for which both Cauchy-Riemann equations are simultaneously satisfied, and hence f'(z) does not exist at any point.
Page 54 #2
(a) f(x+iy) = ix - y + 2 = (2-y) + ix so here u=2-y and v=x. Therefore, u_x = 0 = v_y and u_y = -1 = -v_x. Since these four partial derivatives are continuous and satisfy the Cauchy-Riemann equations, this proves that f'(z) exists everywhere, and that f'(z) = u_x + i v_x = 0 + i. (Note: the fact that the partial derivatives are continuous is a necessary part of the argument and needs to be stated).
To show that f"(z) exists by this method, we write f'(x+iy) = 0 + i so now our u(x,y) = 0 and v(x,y)=i. Therefore, u_x = 0 = v_y and u_y = 0 = -v_x. Since these four partial derivatives are continuous and satisfy the Cauchy-Riemann equations, this proves that (f')'(z) (i.e., f"(z)) exists everywhere, and that f"(z) = u_x + i v_x = 0 + i 0 = 0. (The u and v here are different from those in the previous paragraph, referring now to the real and imaginary parts of f'(z) rather than f(z)).
(Note: this function is one for which f'(z) and f"(z) are obvious from the rules for differentiation, so if it weren't for the fact that you were asked to do this question using the Cauchy-Riemann theory, you wouldn't have needed to do all this).
Parts (b) through (d) are done in exactly the same way; I won't write out all the details.
Page 54 #3
(a) f(x+iy) = 1/(x+iy) = (x-iy)/(x^2 + y^2) so u(x,y) = x/(x^2+y^2) and v(x,y) = -y/(x^2+y^2). A quick calculation gives u_x = (y^2-x^2) / (x^2+y^2)^2 = v_y and u_y = 2xy / (x^2+y^2)^2 = -v_x. Since these four partial derivatives are continuous everywhere except at x=y=0 (where they are undefined), and satisfy the Cauchy-Riemann equations, it follows that f'(z) exists for all non-zero z, and
_
y^2 - x^2 + 2 xyi (x-iy)^2 z 2
f'(z) = u_x + i v_x = ----------------- = - -------- = - ( ----- )
(x^2 + y^2)^2 |z|^4 |z|^2
1 2 2
= - ( - ) = - 1 / z.
z
(b) Here u(x,y) = x^2 and v(x,y)=y^2 so u_x = 2x, u_y=v_x=0, and v_y=2y. Since these four partial derivatives are continuous everywhere, f is differentiable precisely at those points where the Cauchy-Riemann equations are both satisfied. The equation u_y = -v_x is satisfied everywhere and the equation u_x = v_y is satisfied when x=y.
Therefore, f'(z) exists when x=y (i.e., when z is of the form x+ix), and at these points its value is f'(x+ix) = u_x + i v_x = 2x.
(c) f(x+iy) = (x+iy)(y) = xy + iy^2 so u(x,y) = xy and v(x,y)=y^2. Therefore u_x = y, u_y = x, v_x=0, and v_y=2y. Since these four partial derivatives are continuous everywhere, f is differentiable precisely at those points where the Cauchy-Riemann equations are both satisfied. The equation u_x = v_y is satisfied if and only if y=0, and the equation u_y = -v_x is satisfied if and only if x=0.
Therefore, f'(z) exists only at the point z=0. Its value there is f'(0) = u_x(0,0) + i v_x(0,0) = 0 + i 0 = 0.
Page 54 #5
It is only legitimate to write f'(z) = u_x + i v_x when f is differentiable at z, and this happens only if the Cauchy-Riemann equations are satisfied there. Here u(x,y)=x^3 and v(x,y) = (1-y)^3. The Cauchy-Riemann equation u_y=-v_x is satisfied everywhere, but the Cauchy-Riemann equation u_x = v_y is only satisfied when z=i. (To see this, write the equation as 3x^2 = -3(1-y)^2. The LHS is >= 0 while the RHS is <= 0, so the only way they can be equal is if they are both zero, which happens if and only if x=0 and 1-y = 0, i.e, x=0 and y=1, i.e., z=0+1i = i). Therefore, it is only legitimate to write f'(z) = u_x + i v_x when z=i.
Page 54 #6
When z = x+iy is non-zero,
f(x+iy) = ((x-iy)^2)/(x+iy) = ((x-iy)^3)/((x+iy)(x-iy)) = (x^3 - 3ix^2y - 3xy^2 + iy^3)/(x^2+y^2)so u(x,y) = (x^3-3xy^2)/(x^2+y^2) and v(x,y) = (y^3-3yx^2)/(x^2+y^2).
When z=0, u(0,0)=0 and v(0,0)=0.
Because u and v are not defined by a single continuous formula, to find the partial derivatives it is necessary to work straight from the definition.
u(0+h,0) - u(0,0) (h^3-0)/(h^2+0^2) - 0
u (0,0) = lim ----------------- = lim -----------------------
x h->0 h h
h^3
= lim --- = 1
h->0 h^3
u(0,0+h) - u(0,0) (0-0h^2)/(0^2+h^2) - 0
u (0,0) = lim ----------------- = lim ----------------------- = 0
y h->0 h h
v(0+h,0) - v(0,0) (0-0h^2)/(h^2+0^2) - 0
v (0,0) = lim ----------------- = lim ----------------------- = 0
x h->0 h h
v(0,0+h) - v(0,0) (h^3-0)/(0^2+h^2) - 0
v (0,0) = lim ----------------- = lim -----------------------
y h->0 h h
h^3
= lim --- = 1
h->0 h^3
so, at z=0, the Cauchy-Riemann equations are satisfied: u_x = 1 = v_y and u_y = 0 = -v_x.
Page 60 #1
(a) Here u(x,y)=3x+y and v(x,y) = 3y-x. Therefore, u_x = 3 = v_y and u_y = 1 = -v_x. Since all four partial derivatives are continuous and satisfy the Cauchy Riemann equations everywhere, f is differentiable everywhere on the complex plane, and is therefore entire.
(b) Here u(x,y)=sin x cosh y and v(x,y) = cos x sinh y. Therefore, u_x = cos x cosh y = v_y and u_y = sin x sinh y = -v_x. Since all four partial derivatives are continuous and satisfy the Cauchy Riemann equations everywhere, f is differentiable everywhere on the complex plane, and is therefore entire.
(d) Here f(x+iy) = ( (x+iy)^2 - 2) e^(-x) (cos x - i sin y) = e^(-x)[(x^2 - y^2 - 2)(cos x) + 2 x y sin y] + i e^(-x)[(x^2 - y^2 - 2)(sin y) - 2 x y cos x] so u(x,y)= e^(-x)[(x^2 - y^2 - 2)(cos x) + 2 x y sin y] and v(x,y) = e^(-x)[(x^2 - y^2 - 2)(sin y) - 2 x y cos x].
Calculating derivatives shows that u_x = v_y and u_y=-v_x; the calculation is messy but straightforward so I won't write it all out.
Since all four partial derivatives are continuous and satisfy the Cauchy Riemann equations everywhere, f is differentiable everywhere on the complex plane, and is therefore entire.
Note that this is not the best way of showing f is entire! The best way is to write f(z) = (z^2-2) exp(-z) and use the fact that composites, sums, and products of entire functions are entire.
Page 60 #2
(a) Here u(x,y)=xy and v(x,y)=y. The Cauchy-Riemann equation u_x = v_y becomes y = 1 and the equation u_y = -v_x becomes x=0. Therefore, f is differentiable only at the single point z=0+1i = i. Since there is no point with the property that f is differentiable on a complete disk centred at that point, there is no point where f is analytic.
(b) Here u(x,y)=e^y cos x and v(x,y)=e^y sin x. The Cauchy-Riemann equation u_x = v_y becomes -e^y sin x = e^y sin x, satisfied only when sin x = 0 (because e^y is always non-zero), and the equation u_y = -v_x becomes e^y cos x = - e^y cos x, satisfied only when cos x = 0.
Because sin^2 x + cos^2 x = 1, there is no value of x for which both sin x = 0 and cos x = 0. Therefore, there is no point at which f is differentiable, and hence no point at which f is analytic.
Page 60 #4
These are all combinations of polynomials. Polynomials are entire functions (differentiable everywhere). Sums, differences, and products of differentiable functions are entire. Quotients of differentiable functions are differentiable except where the denominator is zero.
Therefore, each of these functions are differentiable except at points where the denominator is zero. In each of these examples, these are just finitely many points (as given in the answers in the book). These are the singular points. The set of complex numbers other than these finitely many exceptions is an open set; f is differentiable everywhere on that open set; therefore, f is analytic everywhere on that open set.
Page 61 #7
(a) Write f(x+iy) = u(x,y) + iv(x,y). Since f is analytic on D, u_x=v_y and u_y=-v_x everywhere on D. Since f is real-valued, v(x,y)=0 and hence v_x = v_y = 0. Therefore, u_x and u_y are also zero, so u is a constant function, and therefore f(x+iy) = u + i0 = u is constant.
(b) Write f(x+iy) = u(x,y) + iv(x,y); then its conjugate is u(x,y) - i v(x,y). Since both of these are analytic, we have u_x = v_y and u_y = -v_x (because f is analytic), and u_x = (-v)_y and u_y = -(-v)_x (because its conjugate is analytic).
Because u_x = v_y = -v_y it follows that u_x = v_y = 0. Because u_y = -v_x = v_x, it follows that u_y = v_x = 0. Since u_x = u_y = 0, u is a constant function. Since v_x = v_y = 0, v is a constant function. Therefore, since f = u + iv, f is constant.
(c) Suppose that |f(z)| is a constant c, and that f is analytic. If c = 0 it follows that f(z)=0 for all z, so f is certainly constant in this case. Now suppose c is non-zero; it then follows that f(z) is non-zero for all z.
Note that
2 2 ---- ____ 2
c = |f(z)| = f(z) f(z) so f(z) = c / f(z).
Since f is analytic and never zero, and the constant function c^2 is
analytic, it follows that the quotient c^2/f(z) is analytic. Therefore
f(z) and
---- f(z)are both analytic. From part (b) it follows that f(z) is constant.
Page 61 #8 (6th edition: page 63 #10)
(a) u_x = 2(1-y), u_(xx) = 0, u_y = -2x, u_(yy)=0, so u_(xx) + u_(yy) = 0 + 0 = 0 proving that u is harmonic.
We seek v such that v_y = u_x = 2(1-y) and v_x = -u_y = 2x. Integrating the second equation with respect to x gives v(x,y) = x^2 + g(y) where g(y) is an arbitrary function of y. Differentiating with respect to y gives v_y = g'(y); plugging that back into the first equation gives g'(y) = 2(1-y) so g(y) = 2y - y^2 + C where C is an arbitrary constant.
This gives v(x,y) = x^2 + 2y - y^2 + C. Since we were just asked for "a" harmonic conjugate of v we can let C=0, giving v(x,y) = x^2 + 2 y - y^2.
(b) u_x = 2 - 3x^2 + 3y^2, u_(xx) = -6x, u_y = 6xy, u_(yy)=6x, so u_(xx) + u_(yy) = -6x + 6x = 0 proving that u is harmonic.
We seek v such that v_y = u_x = 2 - 3x^2 + 3y^2 and v_x = -u_y = -6xy. Integrating the second equation with respect to x gives v(x,y) = -3x^2y + g(y) where g(y) is an arbitrary function of y. Differentiating with respect to y gives v_y = -3x^2 + g'(y); plugging that back into the first equation gives -3x^2 + g'(y) = 2 - 3x^2 + 3y^2 so g'(y) = 2 + 3y^2 and therefore g(y) = 2y + y^3 + C where C is an arbitrary constant. As above, we can take C=0, giving v(x,y) = -3x^2y + 2y + y^3.
As an exercise, you might want to convince yourself that the function u+iv, which the above work shows to be analytic, is in fact the function 2z - z^3.
Page 61 #9 (6th edition: page 63 #11)
(a) If v and V are both harmonic conjugates of u, then v_y = V_y = u_x and v_x = V_x = -u_y. If we let g(x,y)=V(x,y)-v(x,y), we see that g_x = V_x - v_x = 0 and g_y = V_y - v_y = 0, so g is a constant. This proves that the difference V(x,y) - v(x,y) is a constant.
(b) If v is a harmonic conjugate of u, we have u_x = v_y and u_y = -v_x. If u is also a harmonic conjugate of v, we also have v_x = u_y and v_y = -u_x. Combining these, we see that u_x = v_y = -v_y (which means u_x = v_y = 0) and also u_y = v_x = -v_x (which means u_y = v_x = 0).
We therefore have u_x = u_y = 0 everywhere in D, proving that u(x,y) is a constant, and v_x = v_y = 0 everywhere in D, proving that v(x,y) is a constant.