Calculate Limit: \lim_{z\to i} \frac{z^3+i}{z-i}

AdityaDevIf you are trying to solve without using L'Hopital's Rule, then you can use the division algorithm to write$$z^3 + i = (z^2+1)(z-i) + (iz^2-i^2) = (z^2+1)(z-i) -z+1$$and then you can write the limit as$$\lim_{z \to i} \frac{(z^2+1)(z-i)-z+1}{z-i}.$$Now you can divide through by ##z-i## and use the fact that ##z^2+1 = (z+i)(z-i)## to get the limit as$$
  • #1
jjr
51
1

Homework Statement


Calculate the following limit if it exists:

##\lim_{z\to i} = \frac{z^3+i}{z-i}##

Homework Equations


Possibly relevant:
## \lim_{z\to\infty} f(z) = \omega_0 \hspace{5mm} \text{if} \hspace{5mm} \lim_{z\to 0} f\left(\frac{1}{z}\right) = \omega_0##

The Attempt at a Solution


The problem is obviously that the denominator goes to zero, so the solution likely has something to do with rewriting the limit so that this does not happen.

I tried rewriting the first equation to fit the form of the RHS of the possibly relevant equation written above.
##\lim_{z\to i} = \frac{z^3+i}{z-i} = \lim_{z\to 0} \frac{(z^3+i^3)+i}{(z+i)-i} = \omega_0##
so that
## \omega_0 = \lim_{z\to\infty} \frac{(1/z^3 + i^3)+i}{(1/z + i) - i} ##
I still wind up with a zero in the denominator.

I also tried multiplying by the complex conjugate of the expression in the denominator, making the denominator real, but ## \lim_{z\to i} ## implies that the real part goes to zero, so the denominator again goes to zero.

I think I need to factor the expression ## (z^3+i) ## so that I can cancel a term in both the numerator and denominator, but I am having some trouble. The only root I can find is ## z = i ## and I'm not sure how to find the other terms using this information.

Any hints?

Thanks,
J
 
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  • #2
Never mind! Made en error in the polynomial division. Managed to solve it. Disregard thread.
 
  • #3
jjr said:
Never mind! Made en error in the polynomial division. Managed to solve it. Disregard thread.
@jjr, what did you get? When you use polynomial division you get ##z^2 + z + 1 + \frac{2i}{z - i}##, and there is still a problem with the denominator in the remainder fraction.

Edit: My division is incorrect, as pointed out by Dick in a later post...

L'Hopital's Rule might be useful here.
 
Last edited:
  • #4
Mark44 said:
@jjr, what did you get? When you use polynomial division you get ##z^2 + z + 1 + \frac{2i}{z - i}##, and there is still a problem with the denominator in the remainder fraction.

L'Hopital's Rule might be useful here.

##z^3+i=(z-i)(z^2+iz-1)##. Another error in polynomial division??
 
  • #5
Dick said:
##z^3+i=(z-i)(z^2+iz-1)##. Another error in polynomial division??
Oops! You caught me!
 
  • #6
jjr said:

Homework Statement


Calculate the following limit if it exists:

##\lim_{z\to i} = \frac{z^3+i}{z-i}##

Homework Equations


Possibly relevant:
## \lim_{z\to\infty} f(z) = \omega_0 \hspace{5mm} \text{if} \hspace{5mm} \lim_{z\to 0} f\left(\frac{1}{z}\right) = \omega_0##

The Attempt at a Solution


The problem is obviously that the denominator goes to zero, so the solution likely has something to do with rewriting the limit so that this does not happen.

I tried rewriting the first equation to fit the form of the RHS of the possibly relevant equation written above.
##\lim_{z\to i} = \frac{z^3+i}{z-i} = \lim_{z\to 0} \frac{(z^3+i^3)+i}{(z+i)-i} = \omega_0##
so that
## \omega_0 = \lim_{z\to\infty} \frac{(1/z^3 + i^3)+i}{(1/z + i) - i} ##
I still wind up with a zero in the denominator.

I also tried multiplying by the complex conjugate of the expression in the denominator, making the denominator real, but ## \lim_{z\to i} ## implies that the real part goes to zero, so the denominator again goes to zero.

I think I need to factor the expression ## (z^3+i) ## so that I can cancel a term in both the numerator and denominator, but I am having some trouble. The only root I can find is ## z = i ## and I'm not sure how to find the other terms using this information.

Any hints?

Thanks,
J

Expand out and simplify ##(z+i)^3 + i## before taking ##z \to 0##.
 
  • #7
Factoring z3+i: Start by solving [itex] z^{3}=-i=e^{\frac{3\pi}{2}}[/itex], which gives you [itex] z=e^{\frac{i\pi}{2}}=i[/itex], [itex] z=e^{\frac{7\cdot i\pi}{6}}[/itex] and [itex]z=e^{\frac{11\cdot i\pi}{6}} [/itex]. Now dividing by [itex] z-i[/itex] is easy.
 
  • #8
I would like to solve this question.
the numerator tends to 1+i while the denominator tends to 0. so shouldn't the answer be infinite?
 
  • #9
AdityaDev said:
I would like to solve this question.
the numerator tends to 1+i while the denominator tends to 0. so shouldn't the answer be infinite?
Hey, check carefully
Numerator is tending to zero not 1+i
 
  • #10
sorry. i got confused with ##\omega^3## and ##i^3##. Can i directly go for LH rule?
 
  • #11
AdityaDev said:
sorry. i got confused with ##\omega^3## and ##i^3##. Can i directly go for LH rule?
Yeah, sure why not. It's the most easy way. A one step answer.
 
  • #12
AdityaDev said:
I would like to solve this question.
the numerator tends to 1+i while the denominator tends to 0. so shouldn't the answer be infinite?
Sorry. As z→i, z3→-i. As I said above, the quotient is [itex](z-e^{\frac{7\pi i}{6}})\cdot (z-e^{\frac{11\pi i}{6}}) [/itex]. Insert z = i and you are done.
 
  • #13
i wanted to solve without using it.
You can write it as ##\frac{(z+i)(z^2-iz-1)}{z-i}##. I don't know how to proceed.
 
  • #14
Svein said:
Sorry. As z→i, z3→-i. As I said above, the quotient is [itex](z-e^{\frac{7\pi i}{6}})\cdot (z-e^{\frac{11\pi i}{6}}) [/itex]. Insert z = i and you are done.
Can you explain why you took ##e^{\frac{7\pi}{6}}##?
 
  • #15
@AdityaDev
Dick said:
##z^3+i=(z-i)(z^2+iz-1)##.

You get ## \lim_{z\to i} \frac{(z-i)(z^2+iz-1)}{(z-i)} ##, simplify, insert, solved

Edit: Typo
 
  • #16
Edit: Double post
 
  • #17
##a^3+b^3=(a+b)(a^2-ab+b^2)##. so isn't it -iz?
 
  • #18
## (z-i)(z^2+iz-1) = z^3 + iz^2 - z -iz^2 -i^2z+i = z^3+i ##
 
  • #19
jjr said:
## (z-i)(z^2+iz-1) = z^3 + iz^2 - z -iz^2 -i^2z+i = z^3+i ##
But in question the numerator is z3 + i
 
  • #20
Raghav Gupta said:
But in question the numerator is z3 + i
Sorry, made a mistake. Edited now.
 
  • #21
AdityaDev said:
Can you explain why you took e7π6e^{\frac{7\pi}{6}}?
There are three cube roots of -i. In exponential notation, [itex]-i=e^{\frac{3\pi i}{2}} [/itex]. But since [itex]e^{2\pi i}=1 [/itex], we also have [itex]-i=e^{\frac{7\pi i}{2}}=e^{\frac{11\pi i}{2}} [/itex]. Taking the cube root, you divide the exponents by 3, giving the results in post #7.
 
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  • #22
AdityaDev said:
sorry. i got confused with ##\omega^3## and ##i^3##. Can i directly go for LH rule?

AdityaDev said:
sorry. i got confused with ##\omega^3## and ##i^3##. Can i directly go for LH rule?

Even easier: do what I suggested in Post #6: write ##z = i+w## and expand out the numerator.
[tex] \text{numerator} = (i+w)^3 + i = w^3 +3 i w^2 - 3w,[/tex]
so the ratio is ##(w^3 + 3 i w^2 - 3 w)/w =w^2 + 3 i w - 3##. Now taking ##w \to 0## is simple.
 
  • #23
Ray Vickson said:
Even easier: do what I suggested in Post #6: write ##z = i+w## and expand out the numerator.
[tex] \text{numerator} = (i+w)^3 + i = w^3 +3 i w^2 - 3w,[/tex]
so the ratio is ##(w^3 + 3 i w^2 - 3 w)/w =w^2 + 3 i w - 3##. Now taking ##w \to 0## is simple.

What is all this fuss about? ##z-i## is an exact factor of ##z^3+i## by using polynomial division, isn't it? You can just cancel it. Or has the problem changed in some way?
 
  • #24
Dick said:
What is all this fuss about? ##z-i## is an exact factor of ##z^3+i## by using polynomial division, isn't it? You can just cancel it. Or has the problem changed in some way?

No fuss. Some people prefer multiplication to division. Anyway, I have found over and over again that substituting ##x = a + u## and expanding ##f(a+u)## to often be the slickest way to evaluate ##\lim_{z \to a} f(x)##.
 
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FAQ: Calculate Limit: \lim_{z\to i} \frac{z^3+i}{z-i}

What is the concept of limit in mathematics?

The concept of limit in mathematics refers to the value that a function or sequence approaches as the input or independent variable gets closer and closer to a specific value. It is a fundamental concept in calculus and is used to analyze the behavior of functions near a certain point.

How do you calculate a limit?

To calculate a limit, you need to substitute the specific value that the independent variable is approaching into the function and evaluate it. If the resulting value is undefined or indeterminate, further algebraic manipulations or mathematical techniques such as L'Hopital's rule may be used to evaluate the limit.

What is the limit of the given function, \lim_{z\to i} \frac{z^3+i}{z-i}?

The limit of the given function as z approaches i is equal to 2i. This can be determined by substituting i into the function and simplifying the resulting expression.

Why is the limit important in mathematics?

The concept of limit is important in mathematics because it allows us to analyze the behavior of functions near a specific point, which is crucial in understanding the properties and characteristics of various mathematical models and real-world phenomena. It also serves as the foundation for more advanced concepts such as derivatives and integrals in calculus.

What are some real-life applications of limits?

Limits have many real-life applications in fields such as physics, engineering, and economics. For example, in physics, limits are used to analyze the motion of objects and the behavior of physical systems. In economics, limits are used to model and predict the behavior of markets and economies. In engineering, limits are used to design and optimize various systems and structures.

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