Instructions Answer all questions. Give your answers clearly. Do not skip intermediate steps even they are very easy. Complete answers with no error, including expression and notation errors receive full mark. Calculators are not allowed. Cell phones have to be switched off and be kept in your pocked or in your bag. Time 90 minutes. 2 3 4 5 6 7 8 9 0 TOTAL QUESTIONS Q) 4p) Evaluate the following mathematical operations. Write your results in Cartesian form. a) 6e iπ/3 4 + +i 3 b) 2 ) /3 Q2) 9p) A complex function is given as fz) = z 2 +2 z. Replace complex variable z with its Cartesian form z = x+iy in the function and obtain wx,y) = fx+iy). What are the real and imaginary parts ofwx,y). Q3) 9p) A complex function wx,y) = x 2 y 2 +2x+i2xy 2y) is given. Here x and y are real and imaginary parts of complex variablez = x+iy. Check analyticity of the function with Cauchy-Riemann equations. Is it analytic? Q4) 5p) Derive the description of the inverse cosine function w = cos z) in terms of log function. Sami Arıca Page of 9
Hint: Write cosw) in terms of complex exponentials.) Q5) 0p) Evaluate the following integral. 0 20t t 2 + i) 2 dt Hint: change the term in the parenthesis with a new variable.) Q6) 6p) Decide if the following domains are simply connected or not. Write your reasoning? a) b) Q7) 4p) Which of the following loops are positively oriented counter-clockwise direction)? a) y b) y x x Q8) 5p) Evaluate the following line integral. The line is : zt) = Log2)+itπ/3 π/6), 0 t. When it is necessary, take Log2) 0.7 and 3π 5.4. z+)e z dz =? Sami Arıca Page 2 of 9
Hint: d dz zez =? z z =? ora+ib a ib) =?) Q9) 9p) Evaluate the following integral. is a loop with any pointzon it satisfy the inequality z < 3/2. The loop traverses in the counter-clockwise direction. e z z dz =? Hint : Remember Cauchy s integral formula. Q0) 9p) Compute line integral, fz) = 0 2z z 2 2z 3 = z 3 3 z+ C fz) z z 0 dz =?) fz)dz, of the following function Here is a loop with any point z on it satisfy the inequality z < 3. The loop traverses in the counter-clockwise direction. Hint: Remember the integral; z a) n dz =?) C Sami Arıca Page 3 of 9
ANSWERS A) a) 6e iπ/3 = 6 cosπ/3)+i6 sinπ/3) = 3+i3 3 4 +i 3 = 4 i ) 3 +i ) 3 i ) = 3 4 i ) 3 4 = i 3 b) 6e iπ/3 + 4 +i 3 = 3+i3 3+ i 3 = 4+i2 3 = e π+2πk, k Z ) /3 = e π/3+2πk/3, k = 0,,2 2 ) /3 = 2e π/3+2πk/3, k = 0,,2 ) /3 has three roots and therefore 2 ) /3 has three answers. k = 0 : 2e π/3+2πk/3 = 2e π/3 = 2 cosπ/3)+i2 sinπ/3) = +i 3 k = : 2e π/3+2πk/3 = 2e π = 2 cosπ)+i2 sinπ) = 2 k = 2 : 2e π/3+2πk/3 = 2e 5π/3 = 2 cos5π/3)+i2 sin5π/3) = i 3 A2) wx,y) = x+iy) 2 +2x+iy = x 2 +i2xy+iy) 2 +2x iy) = x 2 +i2xy y 2 +2x i2y = x 2 y 2 +2x+i2xy 2y) Re[wx,y)] = x 2 y 2 +2x Im[wx,y)] = 2xy 2y Sami Arıca Page 4 of 9
A3) ux,y) = Re[wx,y)] = x 2 y 2 +2x vx,y) = Im[wx,y)] = 2xy 2y ux,y) = 2x+2 x vx,y) = 2y x ux,y) = 2x y vx,y) = 2x 2 y We check if the the Cauchy-Riemann equations are satisfied. For the given function, x ux,y) y vx,y) y ux,y) = x vx,y) The equality is fulfilled in the second expression but not in the first expression for anyxandy, therefore the function is nowhere analytic. You should immediately realize that the function given in this question is the same with the one given in the second equation. This function contains z in its definition. z is nowhere analytic sofz). Recall the differentiation ofz dz dz dz dz = when dz = dx when we approach to z along real axis) = when dz = idy when we approach tozalong imaginary axis) The derivative should be independent of any direction we approach. Consequently z is nowhere analytic since there isn t any domain it is analytic in the complex plane. Sami Arıca Page 5 of 9
A4) We can write w = cos z) as cosw) = z to escape from the inverse function. Using Euler s formula cosw) is replaced with exponentials. e iw +e iw 2 = z Continue manipulating this equation e iw +e iw e i2w + = 2z = 2ze iw e i2w 2ze iw + = 0 We need to compute roots of the second order equation e iw ) 2 2ze iw + = 0 = z) 2 = z 2 e iw = z)+ = z+ z 2 Then we take logarithm of the both sides of the last equation to extract variablew iw = log z+ ) z 2 w = i log z+ ) z 2 This is the answer A5) Let us changet 2 + i with a new variables. t 2 + i = s 2tdt = ds t = 0 s t = s = i = 2 i Sami Arıca Page 6 of 9
Re-write the integral with the new variable and compute the integral 0 20t t 2 + i) 2 dt = 2 i i = 0 s 0 s 2 ds = 2 i i 2 i i 0 s 2 ds = 0 2 i + 0 i = 0+0i+20 0i 3i = 0 3i = 0+3i) 3i)+3i) = +3i A6) a) Any loop inside the domain stretches a single point in the domain. Equivalently, the interior of any loop in the domain is completely inside the domain. Therefore, this domain is simply connected. b) This loop do not meet the conditions mentioned above. It is not simply connected. A7) a) The direction of the loop is counter-clockwise opposite direction of the clockwise direction). Equivalently, while the loop direction is upward the interior of the loop is at the left side. Therefore direction of the loop is positively oriented. b) The direction of the loop is clockwise. Equivalently, while the loop direction is upward the interior of the loop is at the right side. Therefore direction of the loop is not positively oriented. A8) The functionsz+and e z are entire functions. So, z+)e z is entire. Anti-derivative of the integrandfz) = z+)e z is Fz) = ze z. Then, z+)e z dz = ze z z) z0) = z)e z) z0)e z0) Sami Arıca Page 7 of 9
Here, z0) = Log2)+i0 π/3 π/6) = Log2) iπ/6 z) = Log2)+i π/3 π/6) = Log2)+iπ/6 z0) = z) z0)e z0) = z)e z) = z)e z) Therefore, z)e z) z0)e z0) = i2im [ z)e z)] z)e z) = [ Log2)+iπ/6]e Log2)+iπ/6 We need to compute = Log2)e Log2)+iπ/6 +i π 6 elog2)+iπ/6 = 2Log2)e iπ/6 +i π 3 eiπ/6 Im [ z)e z)] = Im [2Log2)e iπ/6 +i π ] 3 eiπ/6 = Im [ 2Log2)e iπ/6] + Im [i π ] 3 eiπ/6 It follows = 2Log2) sinπ/6)+ π 3 cosπ/6) = Log2)+ π 3 6 z+)e z dz = i2im [ z)e z)] = i = i.4+.8) = i3.2 2Log2)+ π 3 3 ) A9) We will employ Cauchy s integral formula. C fz) z z 0 dz = 2πifz 0 ) Sami Arıca Page 8 of 9
The direction of loopcis counter-clockwise and encloses the pointz 0. fz) is analytic on and at the interior of the loop. e z is entire and direction of the loop is counter-clockwise and the loop encloses z =. Therefore, e z z dz = 2πie = 2πie A0) z = 3 and z = f3) = ±, f ) = ± ) are poles of fz). The loop encloses the poles and its direction is positively oriented. Then, fz)dz = fz)dz+ fz)dz C C 2 with C : r t) = 3+ǫ e i2πt and r 2 t) = +ǫ 2 e i2πt 0 t C fz)dz = C z 3 dz 3 C z+ dz = 2πi 3 0 = 2πi C 2 fz)dz = C 2 z 3 dz 3 C 2 z+ dz = 0 3 2πi = 6πi Concequently, fz)dz = 2πi 6πi = 4πi Sami Arıca Page 9 of 9