MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

sin z is defined and differentiable at all points of C. Hence sin z is entire.
cosh z = ez + e −z
⇒ d 2 dz coshz = ez − e −z
= sinh z
2
cosh z is defined and differentiable at all points of C. So cosh z is entire.
sinh z = ez − e −z
⇒ d 2 dz sinhz = ez + e −z
= cosh z
2
sinh z is defined and differentiable at all points of C. Therefore sinh z is entire.
2. Let z = x + iy. Then we can write
cos z = cos x cosh y − i sin x sinh y
Re(cos z) = cos x cosh y. Let u(x, y) = cos x cosh y and v(x, y) = 0.
u x = − sin x cosh y,
u y = cos x sinh y,
u xx = − cos x cosh y
u yy = cos x cosh y
Then we obtain that ∇ 2 u = u xx + u yy = − cos x cosh y + cos x cosh y = 0.
v xx , v yy = 0 because v(x, y) = 0. So, ∇ 2 v = v xx + v yy = 0.
Hence Re(cos z) is harmonic.
We know that
sin z = sin x cosh y + i cos x sinh y
Then Im(sin z) = cos x sinh y. Let u(x, y) = cos x sinh y and v(x, y) = 0.
u x = − sin x sinh y,
u y = cos x cosh y,
u xx = − cos x sinh y
u yy = cos x sinh y
∇ 2 u = u xx + u yy = − cos x sinh y + cos x sinh y = 0.
Because of v(x, y) = 0, v xx and v yy is zero. Hence ∇ 2 v = v xx + v yy = 0. Im(sin z) is harmonic.
3. We know that cosiz = coshz and siniz = isinhz. Furthermore cos z = cos x cosh y − i sin x sinh y.
cosh z = cos iz = cos(i(x + iy)) = cos(−y + ix) = cos(−y) cosh(x) − i sin(−y) sinh(x)
= cos y cosh x + i sin y sinh x
We know that sin z = sin x cosh y + i cos x sinh y Then,
sinh z =
sin iz
i
= −i sin(iz) = −i sin(−y + ix) = −i(sin(−y) cosh(x) + i cos(−y) sinh(x))
= −i(− sin y cosh x + i cos y sinh x) = cos y sinh x + i sin y cosh x.
4. We know that cos iz = cosh z and sin iz = i sinh z. If we use
cos(z 1 + z 2 ) = cos z 1 cos z 2 − sin z 1 sin z 2
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