- #1

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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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- MHB
- Thread starter jaychay
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You are correct. 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?

- #1

- 58

- 0

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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- #2

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Factorize $(4z^2 + 1) = (4z^2 - i^2)= (2z+i)(2z-i)$

- #3

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How can I solve the limitDaalChawal said:Factorize $(4z^2 + 1) = (4z^2 - i^2)= (2z+i)(2z-i)$

- #4

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MHB

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Substitute the hint DaalChawal gave into the function definition of $f(z)$ and simplify it.jaychay said:How can I solve the limit

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:

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On question 2Klaas 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?

Can you tell me on how to find (a,b) that can make it continue at z=1/2i

- #6

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MHB

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Choose $a$ and $b$ such that $a+bi$ is the same as the limit value from question 1.jaychay said:On question 2

Can you tell me on how to find (a,b) that can make it continue at z=1/2i

Did you find the limit value or are you stuck somewhere?

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- #8

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You made a mistake there bro

$2(\frac{i}{2})+i = 2i$

$2(\frac{i}{2})+i = 2i$

- #9

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MHB

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The first substitution is correct, but the parentheses are placed the wrong way in the substitution of $z=\frac 12 i$.jaychay said:Did I do it correct ?

It should be:

$$\lim_{z\to\frac 12i} f(z) = \lim_{z\to\frac 12i} \frac{4z^2+1}{2z-i} = \ldots = \lim_{z\to\frac 12i} 2z+i = 2\left(\frac 12 i\right) +i=2i$$

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Tha

Thank you broDaalChawal said:You made a mistake there bro

$2(\frac{i}{2})+i = 2i$

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Can you please check me on my another post on complex graph equation problem on question 2 on how to find ( a, b )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?

- #12

- 58

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On question 2Klaas 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:

$$\lim_{z\to\frac 12i} f(z) = \lim_{z\to\frac 12i} \frac{4z^2+1}{2z-i} = \ldots = \lim_{z\to\frac 12i} 2z+i = 2\left(\frac 12 i\right) +i=2i$$

Is ( a,b ) equal to ( 0,2 ) right ?

- #13

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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 2Klaas 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?

Is ( a,b ) equal to ( 0,2 ) right ?

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