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If f(z) = u + iv is an analytic function of z = x + iy and u – v = ex (cosy - siny), then f(z) in terms of z is
- a)e−z2 + (1 + i)c
- b)e- z + (1 + i)c
- c)ez + (1 + i)c
- d)e- 2z + (1 + i)c
Correct answer is option 'C'. Can you explain this answer?
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If f(z) = u + iv is an analytic function of z = x + iy and u – v...
Understanding the Function f(z)
The function f(z) is given as f(z) = u + iv, where u and v are real-valued functions of x and y in the complex plane, defined by the relation u - v = e^x (cos(y) - sin(y)).
Analytic Function Properties
Since f(z) is analytic, it satisfies the Cauchy-Riemann equations:
- ∂u/∂x = ∂v/∂y
- ∂u/∂y = -∂v/∂x
This implies that u and v are interconnected through their partial derivatives.
Finding u and v
Using the provided equation u - v = e^x (cos(y) - sin(y)), we can express v in terms of u:
- v = u - e^x (cos(y) - sin(y))
Now we need to differentiate u and v according to the Cauchy-Riemann conditions and solve for u and v.
Exponential Formulation
Given the structure of e^x and trigonometric functions, we can express the solution in terms of complex exponentials using:
- e^(iy) = cos(y) + i sin(y)
This allows us to rewrite v in an exponential form that connects back to u.
Conclusion: f(z) in Terms of z
After determining the forms of u and v using the Cauchy-Riemann equations and substituting back, we find:
- f(z) = e^z + (1 + i) c
Thus, the correct answer is option 'C': f(z) = e^z + (1 + i) c, which aligns with the analytic properties and the provided relationship of u and v.
The function f(z) is given as f(z) = u + iv, where u and v are real-valued functions of x and y in the complex plane, defined by the relation u - v = e^x (cos(y) - sin(y)).
Analytic Function Properties
Since f(z) is analytic, it satisfies the Cauchy-Riemann equations:
- ∂u/∂x = ∂v/∂y
- ∂u/∂y = -∂v/∂x
This implies that u and v are interconnected through their partial derivatives.
Finding u and v
Using the provided equation u - v = e^x (cos(y) - sin(y)), we can express v in terms of u:
- v = u - e^x (cos(y) - sin(y))
Now we need to differentiate u and v according to the Cauchy-Riemann conditions and solve for u and v.
Exponential Formulation
Given the structure of e^x and trigonometric functions, we can express the solution in terms of complex exponentials using:
- e^(iy) = cos(y) + i sin(y)
This allows us to rewrite v in an exponential form that connects back to u.
Conclusion: f(z) in Terms of z
After determining the forms of u and v using the Cauchy-Riemann equations and substituting back, we find:
- f(z) = e^z + (1 + i) c
Thus, the correct answer is option 'C': f(z) = e^z + (1 + i) c, which aligns with the analytic properties and the provided relationship of u and v.
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If f(z) = u + iv is an analytic function of z = x + iy and u – v...
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 =
dV = ex (sin y + cos y) dx + ez(cosy – siny) dy
∴ dV = d[ex(siny + cosy)]
⇒ 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 =
dV = ex (sin y + cos y) dx + ez(cosy – siny) dy
∴ dV = d[ex(siny + cosy)]
Now,
On integrating
V = ex (siny + cosy) + c1
F(z) = U + iV = ex(cosy - siny) + i ex (siny + cosy) ic1
F = ex(cosy + isiny) + iex (cosy + isiny) + ic1
F(z) = (1 + i) ex + iy + ic1 = (1 + i)ez + ic1
⇒ (1 + i) F(z) = (1 + i) ez + ic1
∴ f(z) = ez + (1 + i) c
On integrating
V = ex (siny + cosy) + c1
F(z) = U + iV = ex(cosy - siny) + i ex (siny + cosy) ic1
F = ex(cosy + isiny) + iex (cosy + isiny) + ic1
F(z) = (1 + i) ex + iy + ic1 = (1 + i)ez + ic1
⇒ (1 + i) F(z) = (1 + i) ez + ic1
∴ f(z) = ez + (1 + i) c
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If f(z) = u + iv is an analytic function of z = x + iy and u – v = ex (cosy - siny), then f(z) in terms of z isa)e−z2+ (1 + i)cb)e- z + (1 + i)cc)ez + (1 + i)cd)e- 2z + (1 + i)cCorrect answer is option 'C'. Can you explain this answer?
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If f(z) = u + iv is an analytic function of z = x + iy and u – v = ex (cosy - siny), then f(z) in terms of z isa)e−z2+ (1 + i)cb)e- z + (1 + i)cc)ez + (1 + i)cd)e- 2z + (1 + i)cCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about If f(z) = u + iv is an analytic function of z = x + iy and u – v = ex (cosy - siny), then f(z) in terms of z isa)e−z2+ (1 + i)cb)e- z + (1 + i)cc)ez + (1 + i)cd)e- 2z + (1 + i)cCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If f(z) = u + iv is an analytic function of z = x + iy and u – v = ex (cosy - siny), then f(z) in terms of z isa)e−z2+ (1 + i)cb)e- z + (1 + i)cc)ez + (1 + i)cd)e- 2z + (1 + i)cCorrect answer is option 'C'. Can you explain this answer?.
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