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The real part of an analytic function f(z) where z = x + iy is given by e-y cos (x). The imaginary part of f(z) is
- a)ey cos (x)
- b)e-y sin (x)
- c)-ey sin (x)
- d)-e-y sin (x)
Correct answer is option 'B'. Can you explain this answer?
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The real part of an analytic function f(z) where z = x + iy is given b...
The real part of an analytic function f(z) is given by e^(-y) * cos(x). We are required to find the imaginary part of f(z).
To find the imaginary part of f(z), we can make use of the Cauchy-Riemann equations. According to these equations, if f(z) is an analytic function, then it satisfies the following conditions:
∂u/∂x = ∂v/∂y (1)
∂u/∂y = -∂v/∂x (2)
where u(x, y) is the real part of f(z) and v(x, y) is the imaginary part of f(z).
Let's differentiate the given real part of f(z) with respect to x and y:
∂u/∂x = -e^(-y) * sin(x) (3)
∂u/∂y = -e^(-y) * cos(x) (4)
Comparing equations (1) and (3), we can see that:
∂v/∂y = -e^(-y) * sin(x) (5)
Comparing equations (2) and (4), we can see that:
∂v/∂x = e^(-y) * cos(x) (6)
Now, integrating equation (5) with respect to y, we get:
v(x, y) = -e^(-y) * sin(x) + g(x) (7)
where g(x) is an arbitrary function of x.
Next, substituting equation (7) into equation (6), we can solve for g(x):
∂v/∂x = e^(-y) * cos(x) (6)
e^(-y) * cos(x) = e^(-y) * cos(x) + g'(x) (8)
g'(x) = 0 (9)
Since g'(x) = 0, it implies that g(x) is a constant.
Therefore, the imaginary part of f(z) is given by:
v(x, y) = -e^(-y) * sin(x) + C (10)
where C is a constant.
Comparing equation (10) with the given options, we can see that the correct answer is option B, i.e., e^(-y) * sin(x).
To find the imaginary part of f(z), we can make use of the Cauchy-Riemann equations. According to these equations, if f(z) is an analytic function, then it satisfies the following conditions:
∂u/∂x = ∂v/∂y (1)
∂u/∂y = -∂v/∂x (2)
where u(x, y) is the real part of f(z) and v(x, y) is the imaginary part of f(z).
Let's differentiate the given real part of f(z) with respect to x and y:
∂u/∂x = -e^(-y) * sin(x) (3)
∂u/∂y = -e^(-y) * cos(x) (4)
Comparing equations (1) and (3), we can see that:
∂v/∂y = -e^(-y) * sin(x) (5)
Comparing equations (2) and (4), we can see that:
∂v/∂x = e^(-y) * cos(x) (6)
Now, integrating equation (5) with respect to y, we get:
v(x, y) = -e^(-y) * sin(x) + g(x) (7)
where g(x) is an arbitrary function of x.
Next, substituting equation (7) into equation (6), we can solve for g(x):
∂v/∂x = e^(-y) * cos(x) (6)
e^(-y) * cos(x) = e^(-y) * cos(x) + g'(x) (8)
g'(x) = 0 (9)
Since g'(x) = 0, it implies that g(x) is a constant.
Therefore, the imaginary part of f(z) is given by:
v(x, y) = -e^(-y) * sin(x) + C (10)
where C is a constant.
Comparing equation (10) with the given options, we can see that the correct answer is option B, i.e., e^(-y) * sin(x).
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Community Answer
The real part of an analytic function f(z) where z = x + iy is given b...
Concept:
If f(z) = u + iv is an analytic function, then it satisfies the following:
Calculation:
Given: u = e-y cos x
∂u/∂x = −e−y sinx
∂u/∂y = −e−y cosx
∂v/∂y = −e−y sinx ---(1)
∂v/∂x = e−y cosx ---(2)
Integrate equation (1) w.r.t. y, taking x as constant, we get:
v = e-y sin x
If f(z) = u + iv is an analytic function, then it satisfies the following:
Calculation:Given: u = e-y cos x
∂u/∂x = −e−y sinx
∂u/∂y = −e−y cosx
∂v/∂y = −e−y sinx ---(1)
∂v/∂x = e−y cosx ---(2)
Integrate equation (1) w.r.t. y, taking x as constant, we get:
v = e-y sin x
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