Question

If f(z) = u + iv is an analytic function of z = x + iy and u – v = e^x (cosy - siny), then f(z) in terms of z is

1. \({e^{ - {z^2}}} + \left( {1 + i} \right)c\)
2. e ^{- z} + (1 + i)c
3. e^z + (1 + i)c
4. e ^{- 2z} + (1 + i)c

Answer 1

Correct Answer - Option 3 : e^z + (1 + i)c

Explanation:

f(z) = u + iv

⇒ i f(z) = - v + i u

⇒ (1 + i) f(z) = (u - v) + i(u + v)

⇒ F(z) = U + iv, where F(z) = (1 + i) f(z)

U = u – v, V = u + v

Now,

Let F(z) be an analytic function

\(dV\; = \;\frac{{ - \partial U}}{{\partial x}}dy\)

dV = e^x (sin y + cos y) dx + e^z(cosy – siny) dy

∴ dV = d[e^x(siny + cosy)]

Now,

On integrating

V = e^x (siny + cosy) + c₁

F(z) = U + iV = e^x(cosy - siny) + i e^x (siny + cosy) ic₁

F = e^x(cosy + isiny) + ie^x (cosy + isiny) + ic₁

F(z) = (1 + i) e^{x + iy} + ic₁ = (1 + i)e^z + ic₁

⇒ (1 + i) F(z) = (1 + i) e^z + ic₁

\(\begin{array}{l} \Rightarrow \;f\left( z \right)\; = \;{e^z} + \frac{i}{{1 + i}}{c_1}\; = \;{e^z} + \frac{{i\left( {1 - i} \right)}}{{\left( {1 + i} \right)\left( {1 - i} \right)}}{c_1}\\ = \;{e^z} + \frac{{\left( {i + 1} \right)}}{2}{c_1} \end{array}\)

∴ f(z) = e^z + (1 + i) c