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Part BCSIR NET December 2024cos-of-i-is-cosh-not-sinh

Cos of i is cosh not sinh

Let f:C→Cf : \mathbb{C} \to \mathbb{C} be the function defined by f(z) = e^((cos(1+i)) sin z). For z = x + iy ∈C\in \mathbb{C}, write f(z) as u(x, y) + iv(x, y), where u, v are real-valued functions. Which of the following is the value of (∂u/∂x)(0,0)(\partial{}u/\partial{}x)(0, 0)?

  1. A.0
  2. B.(e + 1/e)(cos 1)/2✓ Correct
  3. C.(e − 1/e)(cos 1)/2
  4. D.1

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: A harmonic function on any domain has a harmonic conjugate

More on this topic

The chapter behind this: Cauchy–Riemann — what they do and do not give you — free to read

From Analytic Functions › Cauchy–Riemann equations, harmonic functions

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