Aktosun, Spring 2023, Math 5322, Supplementary Problems 2

  1. Determine where the following functions are analytic:
    1.     ez    solution: everywhere
    2.     sinz    solution: everywhere
    3.     cosz    solution: everywhere
    4.     tanz    solution: everywhere except z=π/2+nπ,  n=0,±1,±2,....
    5.     cotz    solution: everywhere except z=nπ,  n=0,±1,± 2,....
    6.     secz    solution: everywhere except z=π/2+nπ,  n=0,±1,± 2,....
    7.     cscz    solution: everywhere except z=nπ,  n=0,±1,± 2,....
    8.     sinhz    solution: everywhere
    9.     coshz    solution: everywhere
    10.     tanhz    solution: everywhere except z=(π/2+nπ)i,  n=0,±1,± 2,....
    11.     sechz    solution: everywhere except z=(π/2+nπ)i,  n=0,±1,± 2,....
    12.     cothz    solution: everywhere except z=nπi,  n=0,±1,± 2,....
    13.     cschz    solution: everywhere except z=nπi,  n=0,±1, ± 2,....
    14.     zn,  n=0,1,2,....    solution: everywhere
    15.     z-n,  n=1,2,3,....    solution: everywhere except at z=0
    16.     principal branch of z1/n,  n=2,3,4,....    solution: everywhere except at z=0
    17.     Logz    solution: everywhere except on the nonpositive real axis.
    18.     rational function    solution: everywhere except at zeros of the denominator
    19.     meromorphic function    solution: everywhere except at the poles
    20.     Argz    solution: nowhere

  2. Express the following in terms of complex logarithms:
    1.     sin-1z    solution: -ilog(iz+(1-z2)1/2)
    2.     cos-1z    solution: -ilog(z+i(1-z2)1/2)
    3.     tan-1z    solution: -(i/2)log((1+iz)/(1-iz))
    4.     cot-1z    solution: -(i/2)log((z+i)/(z-i))
    5.     sec-1z    solution: -ilog(1/z+(1/z2-1)1/2)
    6.     csc-1z    solution: -ilog(i/z+(1-1/z2)1/2)
    7.     sinh-1z    solution: log(z+(1+z2)1/2)
    8.     cosh-1z    solution: log(z+(z2-1)1/2)
    9.     tanh-1z    solution: (1/2)log((1+z)/(1-z))
    10.     coth-1z    solution: (1/2)log((z+1)/(z-1))
    11.     sech-1z    solution: log(1/z+(1/z2-1)1/2)
    12.     csch-1z    solution: log(1/z+(1+1/z2)1/2)

  3. Evaluate the following:
    1.     log(4)    solution: ln(4)+2nπi,  n=0,±1,±2,....
    2.     Log(4)    solution: ln(4)
    3.     log(-1)    solution: (2n+1)πi,  n=0,±1,±2,....
    4.     Log(-1)    solution: πi
    5.     log(i)    solution: (1/2+2ni,  n=0,±1,±2,....
    6.     Log(i)    solution: πi/2
    7.     log(2-3i)    solution: (1/2)ln(13)+(2nπ-a)i, where a=tan-1(3/2) and  n=0,±1,±2,....
    8.     Log(2-3i)    solution: (1/2)ln(13)-αi, where α=tan-1(3/2)
    9.     log((1±i)/√(2))    solution: (±1/4+2ni,  n=0,±1,±2,....
    10.     ii    solution: e(2n-1/2)π,  n=0,±1,±2,....
    11.     (-2)√(2)    solution: e(ln2+(2n+1)πi)√(2),  n=0,±1,±2,....
    12.     2i    solution: eiln2+2nπ,  n=0,±1,±2,....
    13.     1-i    solution: e2nπ,  n=0,±1,±2,....
    14.     ((1±i)/√(2))1+i    solution: e(1+i)(±1/4+2ni,  n=0,±1,±2,....
    15.     (3-4i)1+i    solution: e(1+i)(ln5+(2nπ-a)i, where a=tan-1(4/3) and  n=0,±1,±2,....
    16.     log1+i(1-i)    solution: log(1-i)/log(1+i) or equivalently [(ln2)/2-πi/4+2nπi]/[(ln2)/2+πi/4+2mπi],  n,m=0,±1,±2,....
    17.     1√(2)    solution: e2√(2)nπi,  n=0,±1,±2,....
    18.     e1/2    solution: √(e)
    19.     (e+i0)1/2    solution: ±√(e)
    20.     ei    solution: ei or cos(1)+isin(1)
    21.     (e+i0)i    solution: e(i+2nπ),  n=0,±1,±2,....
    22.     sin-1(1/2)    solution: (2n+1/6)π or (2n+5/6)π,  n=0,±1,±2,....
    23.     cos-1(1/2)    solution: (2n+1/3)π or (2n-1/3)π,  n=0,±1,±2,....
    24.     cos-1(2)    solution: -iln(2±√(3))+2nπ,  n=0,±1,±2,....
    25.     sin-1(i)    solution: -iln(-1+√(2))+2nπ or -iln(1+√(2))+(2n+1)π,  n=0,±1,±2,....
    26.     tanh-1(1+2i)    solution: (1/4)ln(2)+(n+3/8)πi,  n=0,±1,±2,....
    27.     cosh-1(2i)    solution: ln(√(5)+2)+(2n+1/2)πi or ln(√(5)-2)+(2n-1/2)πi,  n=0,±1,±2,....
    28.     tanh-1(1-i)    solution: (1/4)ln(5)+(nπ-π/2+a/2)i, where a=tan-1(2) and  n=0,±1,±2,....

  4. True or false:
    1.     log(z1z2)=log(z1)+log(z2)    solution: true for any nonzero z1, z2
    2.     log(z)+log(z)=2log(z)    solution: not true in general
    3.     log(z)+log(z)+log(z)=3log(z)    solution: not true in general
    4.     log(z-1)=-log(z)    solution: true for any nonzero z
    5.     log(zn)=nlog(z) with n=2,3,....    solution: not true in general
    6.     log(z1/n)=(1/n)log(z) with n=2,3,....    solution: true for any nonzero z
    7.     log(z1/z2)=log(z1)-log(z2)    solution: true for any nonzero z1, z2
    8.     log(1/z)=-log(z)    solution: true for any nonzero z
    9.     Log(z1z2)=Log(z1)+Log(z2)    solution: not true in general, e.g. let z1=z2=-i
    10.     Log(zn)=nLog(z)    solution: true in general only for nonzero z and n=1
    11.     Log(z)+Log(z)=2Log(z)    solution: true for any nonzero z
    12.     Log(z1/z2)=Log(z1)-Log(z2)    solution: not true in general, e.g. let z1=1 and z2=-1
    13.     Log(1/z)=-Log(z)    solution: not true in general, e.g. let z=-1

  5. Characterize the zeros and singularities of the following functions:
    1.     (3z2+z)/(z2+z)    solution: simple zero at z=-1/3, simple pole at z=-1, removable singularity at z=0
    2.     cos2(2z+3i)    solution: double zeros at z=-3i/2+(n/2+1/4)π,  n=0,±1,±2,....
    3.     sech2z    solution: double poles at z=(n+1/2)πi,  n=0,±1,±2,....
    4.     z/(z4+4)    solution: simple zero at z=0, simple poles at z=±√(2)eπi/4 and z=±√(2)ei/4
    5.     tan4z    solution: fourth-order zeros at z=nπ and fourth-order poles at z=(n+1/2)π,  n=0,±1,±2,....
    6.     tan(z2)    solution: simple zeros at z=±√(nπ) and simple poles at z=±√[(n+1/2)π],  n=0,±1,±2,....
    7.     ez/(z2+1)    solution: simple poles at zi
    8.     (sinz)/[z(z2+1)]    solution: simple poles at zi, removable singularity at z=0, simple zeros at z=nπ,  n=±1,±2,±3,....


Tuncay Aktosun aktosun@uta.edu
Last modified: November 28, 2022