Struggling with Complex Number Function? Need Help?

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Discussion Overview

The discussion revolves around solving problems related to complex number functions, specifically focusing on factorization and limits. Participants are seeking assistance with specific questions involving the continuity of functions at certain points and evaluating limits.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Some participants ask for help with factorizing the expression $(4z^2 + 1)$ and propose a factorization as $(2z+i)(2z-i)$.
  • There are multiple requests for guidance on how to solve a limit involving the function $f(z)$ at the point $z=\frac{1}{2}i$.
  • Participants discuss the need to choose values for $a$ and $b$ such that $a+bi$ matches the limit value from a previous question.
  • One participant indicates that a mistake was made in the substitution process, specifically regarding the placement of parentheses in the limit evaluation.
  • Another participant confirms the correctness of the first substitution but reiterates the error in parentheses placement during the limit evaluation.
  • There is a question about whether the values of $(a,b)$ can be equal to $(0,2)$, indicating a search for confirmation on this choice.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of specific substitutions and the factorization process. There is no consensus on the final values of $(a,b)$ or the overall approach to the limit problem, indicating ongoing uncertainty and exploration.

Contextual Notes

Participants reference hints and previous posts, suggesting that there may be additional context or assumptions not fully articulated in the current thread. The discussion reflects a progression of ideas and corrections without resolving all uncertainties.

jaychay
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lim 1.png
lim 2.png


Can you help me with this two questions
I am really struggle on how to do it
Please help me

Thank you in advance
 
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Factorize $(4z^2 + 1) = (4z^2 - i^2)= (2z+i)(2z-i)$
 
DaalChawal said:
Factorize $(4z^2 + 1) = (4z^2 - i^2)= (2z+i)(2z-i)$
How can I solve the limit
 
jaychay said:
How can I solve the limit
Substitute the hint DaalChawal gave into the function definition of $f(z)$ and simplify it.
Then substitute the limit point $z=\frac 12 i$ for $f(z)$ with $z\ne \frac 12 i$ and evaluate it.
How far do you get?
 
Last edited:
Klaas van Aarsen said:
Substitute the hint DaalChawal gave into the function definition of $f(z)$ and simplify it.
Then substitute the limit value $z=\frac 12 i$ for $f(z)$ with $z\ne \frac 12 i$ and evaluate it.
How far do you get?
On question 2
Can you tell me on how to find (a,b) that can make it continue at z=1/2i
 
jaychay said:
On question 2
Can you tell me on how to find (a,b) that can make it continue at z=1/2i
Choose $a$ and $b$ such that $a+bi$ is the same as the limit value from question 1.
Did you find the limit value or are you stuck somewhere?
 
Klaas van Aarsen said:
Choose $a$ and $b$ such that $a+bi$ is the same as the limit value from question 1.
Did you find the limit value or are you stuck somewhere?
[/QUOT
Did I do it correct ?
 

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You made a mistake there bro
$2(\frac{i}{2})+i = 2i$
 
jaychay said:
Did I do it correct ?
The first substitution is correct, but the parentheses are placed the wrong way in the substitution of $z=\frac 12 i$.
It should be:
$$\lim_{z\to\frac 12i} f(z) = \lim_{z\to\frac 12i} \frac{4z^2+1}{2z-i} = \ldots = \lim_{z\to\frac 12i} 2z+i = 2\left(\frac 12 i\right) +i=2i$$
 
  • #10
Tha
DaalChawal said:
You made a mistake there bro
$2(\frac{i}{2})+i = 2i$
Thank you bro
 
  • #11
Klaas van Aarsen said:
Substitute the hint DaalChawal gave into the function definition of $f(z)$ and simplify it.
Then substitute the limit point $z=\frac 12 i$ for $f(z)$ with $z\ne \frac 12 i$ and evaluate it.
How far do you get?
Can you please check me on my another post on complex graph equation problem on question 2 on how to find ( a, b )
 
  • #12
Klaas van Aarsen said:
The first substitution is correct, but the parentheses are placed the wrong way in the substitution of $z=\frac 12 i$.
It should be:
$$\lim_{z\to\frac 12i} f(z) = \lim_{z\to\frac 12i} \frac{4z^2+1}{2z-i} = \ldots = \lim_{z\to\frac 12i} 2z+i = 2\left(\frac 12 i\right) +i=2i$$
On question 2
Is ( a,b ) equal to ( 0,2 ) right ?
 
  • #13
Klaas van Aarsen said:
The first substitution is correct, but the parentheses are placed the wrong way in the substitution of $z=\frac 12 i$.
It should be:
$$\lim_{z\to\frac 12i} f(z) = \lim_{z\to\frac 12i} \frac{4z^2+1}{2z-i} = \ldots = \lim_{z\to\frac 12i} 2z+i = 2\left(\frac 12 i\right) +i=2i$$
Klaas van Aarsen said:
Substitute the hint DaalChawal gave into the function definition of $f(z)$ and simplify it.
Then substitute the limit point $z=\frac 12 i$ for $f(z)$ with $z\ne \frac 12 i$ and evaluate it.
How far do you get?
On question 2
Is ( a,b ) equal to ( 0,2 ) right ?
 

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