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Problem 3: Find the derivative $f'(z)$, where it exists and state where $f$ is analytic. a). $f = \frac{1}{2}x + y + i(\sin x + \cos y)$. b). $f = x^3 + 3xy^2 - 3x + i(y^3 + 3x^2y - 3y)$ c). $f = \frac{1}{z^2 + 3iz - 2}$

Name: problem 3 find the derivative fz where it exists and state where f is analytic a f frac12x y isin x cos y b f x3 3xy2 3x iy3 3x2y 3y c f frac1z2 3iz 2 51135
Uploaded: 2023-05-24T17:04:56-08:00
Duration: 2 min 20 s
Channel: Madhur L
Description: problem 3 find the derivative fz where it exists and state where f is analytic a f frac12x y isin x cos y b f x3 3xy2 3x iy3 3x2y 3y c f frac1z2 3iz 2 51135

          Problem 3:
Find the derivative $f'(z)$, where it exists and state where $f$ is analytic.
a). $f = \frac{1}{2}x + y + i(\sin x + \cos y)$.
b). $f = x^3 + 3xy^2 - 3x + i(y^3 + 3x^2y - 3y)$
c). $f = \frac{1}{z^2 + 3iz - 2}$

$problem 3 find the derivative fz where it exists and state where f is analytic a f frac12x y isin x cos y b f x3 3xy2 3x iy3 3x2y 3y c f frac1z2 3iz 2 51135$

Added by Kathleen P.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

05/24/2023

Step 1

Then $u_x = \frac{1}{2}$, $u_y = 1$, $v_x = \cos x$, $v_y = -\sin y$. The Cauchy-Riemann equations are $u_x = v_y$ and $u_y = -v_x$. $\frac{1}{2} = -\sin y$ and $1 = -\cos x$ have no solution. Therefore, $f'(z)$ does not exist anywhere. $f$ is not analytic Show more…

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Problem 3: Find the derivative $f'(z)$, where it exists and state where $f$ is analytic. a). $f = \frac{1}{2}x + y + i(\sin x + \cos y)$. b). $f = x^3 + 3xy^2 - 3x + i(y^3 + 3x^2y - 3y)$ c). $f = \frac{1}{z^2 + 3iz - 2}$