Sample Solutions to Practice Exam 1 for Dr. Z. Math 403(2) (Rutgers (NB)), Spring 2020 course NAME: Donald Trump Email: donald@whitehouse.org DECLARATION: I HEREBY DECLARE THAT I DID NOT GET ANY HELP FROM A PERSON OR A SMART MACHINE, EXCEPT FOR CONSULTING THE CLASS TEXTBOOK, and the CLASS's On-Line material given at http://sites.math.rutgers.edu/~zeilberg/math403_20.html VERSION OF May 2: Correcting a computational error in #4 (thanks to Dhanvin Patel) ------------------------ 1. By plugging in the values and simplifying yoy get 1/Z^2=-i. So Z^2=i=exp(I*Pi/2), exp(5*Pi/2) so Z=1*exp(I*Pi/4),1*exp(I*5*Pi/4)= cos(Pi/4)+I*sin(Pi/4), cos(5*Pi/4)+I*sin(5*Pi/4) So Z=sqrt(2)/2+I*sqrt(2)/2 or Z=-sqrt(2)/2-I*sqrt(2)/2 . Ans. to 1: sqrt(2)/2+I*sqrt(2), -sqrt(2)/2-I*sqrt(2)/2 -------------------------- 2. The only poles are z=1 and z=I. The residue theorem says that the contour integral over a closed curve Gamma is 2*Pi*I times the sum of the residues of the poles INSIDE GAMMA. (a): Neither poles are inside |z-1-I|=1/5 so the integral is 0 (also follows from Cauchy's theorem) (b): Both poles are inside |z|=5. The residue at z=1 is 1^9/(1-I)=1/(1-I). The residue at z=I is I^9/(I-1)=-I/(1-I) whose sum is 1, hence the integral is 2*Pi*I (c): The only pole inside |z-i|=1/2 is z=i, the residue is i^9/(i-1)=-i/(1-i)=-i*(1+i)/|1+i|^2= (1-i)/2. Hence the contour integral is 2*Pi*i*((1-i)/2)= Pi (1+i) (d): The only pole inside |z-1|=1/2 is z=1, the residue is 1^9/(1-i)=(1+i)/2 Hence the contour integral is 2*Pi*i*((1+i)/2)= Pi (-1+i) Ans. to 2): a)0 b)2*Pi*i, c) Pi+ i*Pi (d) -Pi+ i*Pi -------------------------- Sol. to 3. We have u=3*x^6+3*y, v=11*x+7*y^2. v_x=11 , u_y=3, hence the integrand is v_x - u_y = 8. The region is a rectangle of area (120-100)*(1003-1000)=60. Since the area integral over any region of a constant is the area of that region times that constant, the area integral, and hence the desired line integral when the diretion is the default direction (counterclockwise) is 8*60=480. But we are told that the direction of motion is CLOCKWISE, so we have to multiply by -1. Ans. to 3: -480 . -------------------------- Sol. to 4: The Taylor polynomial of degree 3 of sin(z) is z- z^3/6 Plugging-in z=i/2 gives i/2 - (1/6)*(i/2)^3=i/2+ i/48=25/48*i Ans. to 4: (25/48)*i -------------------------- Sol. to 5: The Taylor polynomial of degree 2 of f(z) at z=z0 is f(z0)+ f'(z0)*(z-z0)+ f''(z0)/2*(z-z0)^2 Here z0=i and we have f(z)=z^4, f'(z)=4*z^3, f''(z)=12*z^2 f(i)=i^4=1, f'(i)=4*i^3=-4i , f''(i)=12*i^2=-12 Hence The Taylor polynomial of degree 2 of f(z)=z^4 at z=i is 1 -4i*(z-i)-6*(z-i)^2 Ans. to 5: 1 -4i*(z-i)-6*(z-i)^2 -------------------------- Sol. to 6: Recall that the change of argument travelling COUNTER-CLOCKWISE along a closed contour of a function with at worst poles is the number of Zeros MINUS the number of poles INSIDE the countour. This is only valid if none of the zeros and none of the poles lie ON the contour. Note that f(z) has zeros at z=i, z=2i, z=3i and a pole at z=10i Sol. to a): None of the zeros and none of the poles lie on the contour. All three zeros lie inside the semi-circle, but the pole does not. Hence the required difference is 3-0=3. Multiplying by 2*Pi, we get that the desired change of argument 6*Pi . Sol. to b): None of the zeros and none of the poles lie on the contour. All three zeros lie inside the semi-circle, and so does the pole z=10i. Hence the required difference is 3-1=2. Multiplying by 2*Pi, we get that the desired change of argument 4*Pi . BUT, now we are told that the trip goes clockwise, so we have to multiply by minus one, getting -4*Pi Ans. to 6: a) 6*Pi b) -4*Pi -------------------------- Sol. to 7: We use the Cauchy-Riemman equations: u_x=v_y, u_y=-v_x . For a): u=2x, v=-3y, so u_x=2, v_y=-3 , these are NOT equal, so it is NOT analytic For b): u= y, v=-x, so u_x=0, v_y=0, 0=0, OK!, also u_y=1, v_x=-1 1=-(-1) also OK! Analytic For c): u= x^2-y^2, v=2*x*y, so u_x=2x, v_y=2x, 2x=2x, OK!, also u_y=-2y, v_x=2y, -2y=-(2y) also OK! Analytic For d): u=x^2+y^2, v=2xy so u_x=2x, v_y=2x , so far so good. Also u_y=2y, v_x=2y, So it is NOT true that u_y=-v_x so this function flunks the second Cauchy-Riemann equatio, so Analytic Ans. to 7: {b,c} -------------------------- Sol. to 8: We compare f(z)=z^5-i to -i on |z|=0.99 |z^5-i+i|=|z^5|=|z|^5=(0.99)^5<|i| so the number of zeros of f(z) in |z|<0.99 is the same as the number of zeros of -i, namely NO zeros. We compare f(z)=z^5-i to z^5 on |z|=1.01 |z^5-i-z^5|=|-i|=1<|-z^5|=1 <1.01^5 so the number of zeros of f(z) in |z|<1.01 is the same as the number of zeros of z^5 in |z|<1.01, namely 5 (counting multiplicity, of course). By the Fundamental Theorem of Algebra (or for that matter, in this simple case, directly) f(z) has altogether five zeros. None of them lie in |z|<0.99, hence all of them lie in the annulus 0.99<|z|<1.01 Comment: We can replace 0.99 by any positive real number less than 1 and replace 1.01 by any positive real number large than 1, no matter how close, so the above reasoning implies that the five zeros of f(z)=z^5-i all lie ON |z|=1. Of course, this can be done directly w/o Rouche, just solve z^5=i, like we did in Lecture 2, but the point of this problem is to test your knowledge of Rouche's theorem, and that you can apply it correctly. ------------------------------------------------------------------------------ Sol. to 9: a) This function has neither zeros nor poles, so the change of argument over ANY closed contour is ALWAYS ZERO. WRONG! b) The change of argument is always an INTEGER (either positive or negative) times 2*Pi. 3*Pi/(2*Pi)=3/2. WRONG! c) f(z)=1/(z^9+1) has NO zeros, but 9 poles INSIDE |z|=5, hence the change of argument as it travells the usual (default orienation) if 2*Pi(0-9)=-18*Pi. But here we are travelling CLOCKWISE, to the desired answer is 18*Pi, so it is CORRECt. d) f(z)=tan(z)=sin(z)/cos(z) has ONE zero, namely z=0 in |z|<1 and NO poles (the closest pole is z=Pi/2=1.57..). So the change of argument is 2*Pi*(1-0)=2*Pi. CORRECT. Ans. to 9: The following are wrong: {a,b} ------------------------------------------------------------------------------ Sol. to 10: Such a function DOES NOT EXIST. If the domain is OPEN, then the range under an analytic function on that domain must be also OPEN. {z: |z|<1} is OPEN while {z: |z-3|<=2} is CLOSED, so NO WAY. ------------------------------------------------------------------------------ Sol. to 11: This is a special case of the MEAN VALUE THEOREM with z0=2i, f(z)=z^9 and r=10. f(z) is ALWAYS analytic, so the value of the integral equals 2*Pi*f(2i)= 2*Pi*(2i)^9=2^10*Pi*i= 1024*Pi*i Ans. to 11: 1024*Pi*i ------------------------------------------------------------------------------ Sol. to 12: Consider g(z)=f(10z)/100. g(z) is analytic in |z|<1 , g(0)=0 and |g(z)|<=1. By Rouche, we have the stronger inequality |g(z)|<=|z|. By the second part of Rouche, it if so happens that there exists z0 such that |z0|<1 and |g(z0)|=|z0| then g(z)=cz for some constant c (with, of course |c|=1). Here |g(1/2*i)|=|30+40i|/100=1/2, so g(z)=cz, where g(1/2*i)=(3+4i)/10 so c= ((3+4i)/10)/(1/2*i)=-4/5+3/5*i and g(z)= (-4/5+ i*3/5)*z, hence f(z)=100*g(z/10)= (-4/5+ i*3/5)*(10*z)= (-8+6i)*z Ans. to 12: f(z) must be given by f(z)=(-8+6i)*z COMMENT: One does not have to change to g(z). It follows that f(z)=cz for SOME constant, and one can directly use the fact that f(5i)= 30+40*i to find the constant. ------------------------------------------------------------------------------ Sol. to 13: For any analytic function the integral of f(z)/(z-z0)^(k+1) over any contour GAMMA containing z0 is f^(k)(z0)/k! TIMES 2*Pi*i. Here f(z)=e^(z^2) z0=i, k=1. By the chain rule: f'(z)=(2z)*e^(z^2) so f'(i)=2*i*e^(i^2)= 2*i*exp(-1) Multiplying by 2*Pi*i, we get 2*i*exp(-1) *(2*Pi*i)=-4*Pi/e . Ans. to 13: -4*Pi/e . ------------------------------------------