analytic function

logistic_guy

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here is the question

Determine the singular points of the function and state why the function is analytic everywhere else: \(\displaystyle f(z) = \frac{z^3 + i}{z^2 - 3z + 2}\).


my attemb
i think when i've fraction, i don't want denomator to be zero
\(\displaystyle z^2 - 3z + 2 = (z - 2)(z - 1) = 0\)
so \(\displaystyle z = 1\) and \(\displaystyle z = 2\)
is there a reason to ask me why the function is analytic everywhere else?:rolleyes:
 
here is the question

Determine the singular points of the function and state why the function is analytic everywhere else: \(\displaystyle f(z) = \frac{z^3 + i}{z^2 - 3z + 2}\).


my attemb
i think when i've fraction, i don't want denomator to be zero
\(\displaystyle z^2 - 3z + 2 = (z - 2)(z - 1) = 0\)
so \(\displaystyle z = 1\) and \(\displaystyle z = 2\)
is there a reason to ask me why the function is analytic everywhere else?:rolleyes:
You've found where the denominator has a problem. Are there any problems in the numerator?

-Dan
 
here is the question

Determine the singular points of the function and state why the function is analytic everywhere else: \(\displaystyle f(z) = \frac{z^3 + i}{z^2 - 3z + 2}\).


my attemb
i think when i've fraction, i don't want denomator to be zero
\(\displaystyle z^2 - 3z + 2 = (z - 2)(z - 1) = 0\)
so \(\displaystyle z = 1\) and \(\displaystyle z = 2\)
is there a reason to ask me why the function is analytic everywhere else?:rolleyes:
A function f(z) is said to be analytic in a region R of the complex plane if f(z) has a derivative at each point of R and if f(z) is single valued.
 
\(\displaystyle f(z) = \frac{z^3 + i}{z^2 - 3z + 2}\)

\(\displaystyle f'(z) = \frac{3z^2(z^2 - 3z + 2) - (2z - 3)(z^3 + i)}{(z^2 - 3z + 2)^2}\)

Do ALL the derivatives exist & continuous everywhere within the region?
Please share your work!
yes it is exist & continuous everywhere except when \(\displaystyle (z^2 - 3z + 2)^2 = 0\)
\(\displaystyle z^2 - 3z + 2 = 0\)
\(\displaystyle (z - 2)(z - 1) = 0\)
\(\displaystyle z - 2 = 0\) and \(\displaystyle z - 1 = 0\)
\(\displaystyle z = 2\) and \(\displaystyle z = 1\)

remove this two points discontinuties and i guarantee to you the function is continuous everywhere without even touching cauchy riemann equations

but i want to learn how to apply cauchy riemann equations to this function☹️
i don't know how. can you show me how to do it?😭
 
\(\displaystyle f(z) = \frac{z^3 + i}{z^2 - 3z + 2}\)

\(\displaystyle f'(z) = \frac{3z^2(z^2 - 3z + 2) - (2z - 3)(z^3 + i)}{(z^2 - 3z + 2)^2}\)


yes it is exist & continuous everywhere except when \(\displaystyle (z^2 - 3z + 2)^2 = 0\)
\(\displaystyle z^2 - 3z + 2 = 0\)
\(\displaystyle (z - 2)(z - 1) = 0\)
\(\displaystyle z - 2 = 0\) and \(\displaystyle z - 1 = 0\)
\(\displaystyle z = 2\) and \(\displaystyle z = 1\)

remove this two points discontinuties and i guarantee to you the function is continuous everywhere without even touching cauchy riemann equations

but i want to learn how to apply cauchy riemann equations to this function☹️
i don't know how. can you show me how to do it?😭
You mean, you need a hint that you could look up yourself?

-Dan
 
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