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

[jjr]

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

 

[jjr]

Never mind! Made en error in the polynomial division. Managed to solve it. Disregard thread.

 

[Mark44]

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.

 

[Dick]

@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??

 

[Mark44]

$z^3+i=(z-i)(z^2+iz-1)$. Another error in polynomial division??

Oops! You caught me!

 

[Ray Vickson]

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 $z \to 0$.

 

[Svein]

Factoring z³+i: Start by solving $z^{3}=-i=e^{\frac{3\pi}{2}}$, which gives you $z=e^{\frac{i\pi}{2}}=i$, $z=e^{\frac{7\cdot i\pi}{6}}$ and $z=e^{\frac{11\cdot i\pi}{6}}$. Now dividing by $z-i$ is easy.

 

[AdityaDev]

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?

 

[Raghav Gupta]

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

 

[AdityaDev]

sorry. i got confused with $\omega^3$ and $i^3$. Can i directly go for LH rule?

 

[Raghav Gupta]

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.

 

[Svein]

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 $(z-e^{\frac{7\pi i}{6}})\cdot (z-e^{\frac{11\pi i}{6}})$. Insert z = i and you are done.

 

[AdityaDev]

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.

 

[AdityaDev]

Sorry. As z→i, z³→-i. As I said above, the quotient is $(z-e^{\frac{7\pi i}{6}})\cdot (z-e^{\frac{11\pi i}{6}})$. Insert z = i and you are done.

Can you explain why you took $e^{\frac{7\pi}{6}}$?

 

[jjr]

@AdityaDev

$z^3+i=(z-i)(z^2+iz-1)$.

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

Edit: Typo

 

[jjr]

Edit: Double post

 

[AdityaDev]

$a^3+b^3=(a+b)(a^2-ab+b^2)$. so isn't it -iz?

 

[jjr]

$(z-i)(z^2+iz-1) = z^3 + iz^2 - z -iz^2 -i^2z+i = z^3+i$

 

[Raghav Gupta]

$(z-i)(z^2+iz-1) = z^3 + iz^2 - z -iz^2 -i^2z+i = z^3+i$

But in question the numerator is z³ + i

 

[jjr]

But in question the numerator is z³ + i

Sorry, made a mistake. Edited now.

 

[Svein]

Can you explain why you took e7π6e^{\frac{7\pi}{6}}?

There are three cube roots of -i. In exponential notation, $-i=e^{\frac{3\pi i}{2}}$. But since $e^{2\pi i}=1$, we also have $-i=e^{\frac{7\pi i}{2}}=e^{\frac{11\pi i}{2}}$. Taking the cube root, you divide the exponents by 3, giving the results in post #7.

 

[Ray Vickson]

sorry. i got confused with $\omega^3$ and $i^3$. Can i directly go for LH rule?

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.
$$\text{numerator} = (i+w)^3 + i = w^3 +3 i w^2 - 3w,$$
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.

 

[Dick]

Even easier: do what I suggested in Post #6: write $z = i+w$ and expand out the numerator.
$$\text{numerator} = (i+w)^3 + i = w^3 +3 i w^2 - 3w,$$
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?

 

[Ray Vickson]

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 $\lim_{z \to a} f(x)$.