MATH 417 Homework 4 Instructor: D. Cabrera Due July 7. z c = e c log z (1 i) i = e i log(1 i) i log(1 i) = 4 + 2kπ + i ln ) cosz = eiz + e iz

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1 MATH 47 Homework 4 Instructor: D. abrera Due July 7. Find all values of each expression below. a) i) i b) cos i) c) sin ) Solution: a) Here we use the formula z c = e c log z i) i = e i log i) The modulus of i is r = and the principal argument is Θ = π 4. Therefore, log i) = ln + i π ) 4 + kπ, k = 0, ±, ±,... Multiplying log i) by i we get i log i) = i ln + i π )] 4 + kπ π ) i log i) = 4 + kπ + i ln ) Finally, we exponentiate i log i) to get where k = 0, ±, ±,.... b) Here we can use either i) i = e π/4+kπ+i ln i) i = e π/+kπ e i ln i) i = e π/4+kπ cos ln ) + i sin ln )] cosz = eiz + e iz cos i) = ei i) + e i i) cos i) = e+i + e i

2 or c) Here we use the formula cos z = cosh x cosy i sinh x sin y cos i) = cosh cos ) i sinh sin ) cos i) = cosh cos + i sinh sin sin z = i log iz + z ) /] sin = i log i ± ] 3i sin = i log ± ] 3)i If we take the positive root, then we have sin = i log + ] 3)i sin = i ln + π )] 3) + i + kπ sin = π + kπ i ln + 3) If we take the negative root, then we have sin = i log ] 3)i sin = i ln π )] 3) + i + kπ where k = 0, ±, ±,.... sin = π + kπ i ln 3). Prove that sinz) = sin z cos z by using the definitions of sin z and cos z. Solution: Using the definition of sin z we have sinz) = eiz) e iz) i sinz) = eiz e iz )e iz + e iz ) i ) e iz e iz e iz + e iz sinz) = i sinz) = sin z cosz where in the last step, we used the definitions of sin z and cosz. )

3 3. Find the values of z for which cosz = 0 by using the fact that cosz = cos x + sinh y where sinh y = ey e y Solution: If cosz = 0 then cosz = 0. So it must be the case that both cos x = 0 and sinh y = 0 happen simultaneously. From the first equation we have k + )π x =, k = 0, ±, ±,... From the second equation we have y = 0. Therefore, cosz = 0 when z = k + )π, k = 0, ±, ±, Show that fz) = sin z) is analytic nowhere. Solution: The function can be written as sin z) = sinx iy) sin z) = sin x cosh y) + i cosxsinh y) sin z) = sin x cosh y i cosxsinh y Letting u = sin x cosh y and v = cos x sinh y and computing their first partial derivatives we get u x = cosxcosh y, u y = sin x sinh y, v y = cosxcosh y v x = sin x sinh y In order for the auchy-riemann equations u x = v y, u y = v x ) to be satisfied, we need cosxcosh y = 0 and sin x sinh y = 0 to occur simultaneously. From the first equation we can only have cosx = 0 since cosh y > 0 for all y. Therefore, k + )π x =, k = 0, ±, ±,... From the second equation we must have sinh y = 0 because sin x and cos x cannot be 0 simultaneously. Therefore, y = 0. Thus, since the first partial derivatives of u and v are continuous everywhere in the k + )π complex plane and the auchy-riemann equations are satisfied for z =, f z) exists for these values of z. However, at each point there is no neighborhood throughout which fz) is analytic. Therefore, fz) = sin z) is analytic nowhere.

4 5. Evaluate the integral e z dz where is the contour consisting of the two straight-line segments: ) from z = i to z = + i and ) from z = + i to z = i. Solution: To evaluate the integral we integrate over each line segment and then add the results. On the first segment we have the parametrization zt) = t + i, 0 t Therefore, the integral of fz) over this segment is fz) dz = = 0 0 fzt))z t) dt e t+i ) dt = e t+i 0 = e +i e i On the second segment we have the parametrization zt) = + it, t Therefore, the integral of fz) over this segment is fz) dz = The value of the integral is then fz) dz = fz) dz + fz) dz fz) dz = e +i e i + e i e +i = fzt))z t) dt e +it i) dt = e +it = e i e +i = e i e i = ecos i sin ) + ecos + i sin ) = e cos + cos ) + i sin sin )]

5 Note: Instead of using parametrizations, we could have said that fz) is entire so it has an antiderivative Fz) = e z and the value of the integral is e z dz = F i) Fi) = e i e i which is exactly what we obtained above. 6. Evaluate the integral z ) dz where is the semicircle z = e it, π t π Solution: The value of the integral is b fz) dz = fzt))z t) dt z ) dz = = i a π/ π/ π/ π/ e it ) ie it dt e 3it e it) dt ] π/ = i 3i e3it i eit π/ ) = 3 ei3π/) e iπ/) = 3 i i 3 i i oriented counterclockwise. ) 3 ei 3π/) e i π/) = 8 3 i Note: Instead of using the parametrization, we could have said that fz) is entire so it has an antiderivative Fz) = 3 z3 z and the value of the integral is z ) dz = Fi) F i) = which is exactly what we obtained above. ) ) 3 i3 i 3 i)3 i) = 3 i i 3 i i = 8 3 i 7. Show that z + z 4 dz π

6 where is the upper half of the circle z = oriented counterclockwise. Justify your answer. Solution: The length of the contour is L = π. Now we must find an upper bound on fz). Using the triangle inequality z + z z + z on the numerator we have z + z + = + = 3 Using the triangle inequality z z z z on the denominator we have z 4 z 4 = 4 = 3 Thus, the modulus of fz) satisfies the inequality fz) = z + z = hoosing M = and using the formula for the ML-Bound we have z + z 4 dz ML = π 8. Find an upper bound on dz z + where is the circle z i = oriented counterclockwise. Justify your answer. Solution: The length of the contour is L = π. To find an upper bound on fz) we ll factor z + and take the modulus to get z + = z i z + i = z i) + i = z i) + i Now we use the triangle inequality z + z z z on the denominator to get Thus, we have z i) + i z i i = = z + = hoosing M = and using the ML-Bound formula we have dz z + ML = π