MATH 106 HOMEWORK 4 SOLUTIONS. sin(2z) = 2 sin z cos z. (e zi + e zi ) 2. = 2 (ezi e zi )

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1 MATH 16 HOMEWORK 4 SOLUTIONS 1 Show directly from the definition that sin(z) = ezi e zi i sin(z) = sin z cos z = (ezi e zi ) i (e zi + e zi ) = sin z cos z Write the following complex numbers in standard form: (i) ( 1 + i 3) i What is the principal value? (ii) tan 1 (i), (iii) tan( iπ ), (iv) solve the equation sin z = (i) ( 1+i 3) i = e i(ln()+i( π 3 +nπ)) = e ( π 3 +nπ) cos(ln )+ie ( π 3 +nπ) sin(ln ) (ii) (iii) (iv) PV ( 1 + i 3) i = e π 3 cos(ln ) + ie π 3 sin(ln ) tan 1 (i) = z tan(z) = i eiz e iz e iz = + e iz e iz e iz = e iz e iz e iz = 3e iz e iz = 3 z = π + i ln 3 + nπ tan( iπ iπ ei( ) e i( iπ ) ) = i(e i iπ + e i( iπ ) ) = i e π e π e π + e π sin z = e iz e iz = 4i w 1 w = 4i, with w = eiz w 4iw 1 = w = ( ± 3)i e iz = ( ± 3)e iπ z = π + kπ + i ln( ± 3) 1

2 3 Evaluate the following integrals by parametrizing the contour: (i) xdz where is the oriented line segment joining 1 to i, (ii) (z 1)dz where is the semicircle joining to, (iii) cos( z )dz where is the line segment joining to π + i (i) Parametrize the contour by z = (i 1)t + 1 with t 1 Then 1 xdz = (1 t)(i 1)dt = i 1 (ii) Using the parametrization we compute (z 1)dz = z = 1 + e iθ, θ, dz = ie iθ dθ, e iθ ( ie iθ )dθ = i e iθ dθ = 1 θ= e iθ = θ= (iii) Writing z = (π + i)t for t 1, we compute cos( z π + i )dz = 1 4 Evaluate e i( π +i)t +e i( π +i)t dt = i (e i( π +i)t e i( π +i)t) t=1 = t= = e + e 1 z 1+i dz where is the positively oriented unit circle, and the integrand is defined by chosing the branch < arg(z) < π What happens if we take < arg(z) < π? For z = e iθ, we have π z 1+i dz = e iθ(i 1) ie iθ dθ = i π If we take < arg(z) < π we ll get i(e π e ) 5 ompute z a dz e θ dθ = i(1 e ) where is the counterclockwise unit circle and a is any real number The principal value is used for the integrand Do it in two ways: by picking a parametrization of, and by using a suitable anti-derivative

3 Let z = e iθ If a 1, we have π z a dz = e iaθ ie iθ dθ = eiπ(a+1) e iπ(a+1) If a = 1, then z 1 dz = π idθ = πi = i sin(()π) For the second method, let us assume first a 1 Note that the principal value of z a = e alog z is undefined at the negative reals To compute the integral, we will split the circle into the upper and lower halves For the integral across the upper half 1, we will replace z a by a different branch which is everywhere defined and holomorphic We pick a branch cut which doesn t cross 1, for instance π < arg(z) < 3π This branch of z a agrees with the principal branch along 1 Then, z 1 a dz = e(a+1) log(z= 1) e (a+1) log(z=1) = eπi(a+1) 1 For the integral along, we may branch cut along a half-line which avoids, for instance (1) 3π < arg(z) < π The values of z a for this branch coincide with the principal values along We evaluate z a dz = e(a+1) log(z=1) e (a+1) log(z= 1) Putting things together z a dz = eπi(a+1) e i(a+1) = 1 ei(a+1) just as before The case a = 1 is done in similar way but using the antiderivative of 1 z which is Log (z) and was done in class NB You should be careful when chosing your branch cut It is especially easy to get a wrong answer when evaluating the integral along Picking a branch cut which avoids is not enough, you have to make sure that the chosen branch coincides with the principal value, so that you are not changing the integral For instance the branch cut π () < arg(z) < 5π would not work in this case This is because this branch of z a does not agree with the principal value along This can be seen, for instance, at the 3

4 4 point 1 There, the branch () gives the value 1 a+1 = e πi(a+1) which differs from the principal value 1 a+1 = 1 However, you can convince yourself that the chosen branch (1) does work 6 Show that 1 1 z i dz = (1 + e )(1 i) for any path joining 1 to 1 which lies above the real axis, endpoints excluded The principal value is used for the integrand Do it also for any path joining i and i which lies on the right of the imaginary axis Observe that the integrand does not exist at the endpoint z = 1 To fix this, let us consider a different branch cut along π < arg(z) < 3π The branch of z i above agrees along the path of integration with the principal branch (except possibly at z = 1 where the latter is undefined) Therefore, we may safely work with the new branch considered above Now, note that z i has an antiderivative in the upper half plane given by z i+1 where the new branch is used again in the definition of z i+1 Therefore 1 1 z i dz = e(i+1) log(z=1) e (i+1) log(z= 1) = 1 e(i+1)(iπ) = (1 + e )(1 i) The integral along the second path causes no problems since the path does not intersect the branch cut at the negative reals Therefore, i i z i dz = e(i+1) log(z=i) e (i+1) log(z= i) = (e π + e π )() 7 What are the values of the following integrals: z z 3 = iπ iπ (i+1) e(i+1) e (i) dz where is the positively oriented unit circle, (ii) Log(z + )dz where is the positively oriented unit circle In both cases the integrands are holomorphic on and inside the unit circle so by auchy Theorem the integrals are For the first function, the pole is at z = 3 which is clearly outside The second function is holomorphic everywhere except for the line z = +x with x is a negative real This line also avoids the unit circle

5 8 Determine the value of the integral (z 1) n dz where n is any integer and is a positively oriented square of side a, which doesn t go through 1 If 1 is not inside the square, the integral is because (z 1) n is holomorphic inside no matter whether n is positive, negative or If 1 is inside the square, and n the same reasoning shows that the integral is So let us consider the case when n <, in which case the function (z 1) n has a pole at z = 1 We can consider a small circle around 1 that lies inside the square Since (z 1) n is holomorphic in the area between the square and the circle, the integrals over these curves are equal onsider parametrization of the circle z = 1 + re iθ, where r is the radius Then π (z 1) n dz = r n+1 e inθ ie iθ dθ = irn+1 θ=π n + 1 ei(n+1)θ = The case n = 1 is special since then the denominator becomes The integral can be computed by hand to be πi 9 Show that the area enclosed by a positively oriented simple closed curve is given by 1 zdz i θ= 5 We have zdz = (x iy) (dx + idy) = (x dx + y dy) + i(x dy y dx) Let R be the region encolosed by We can apply Green s theorem to each of the two terms above to conclude zdz = i dx dy = i area (R) R