Calculating the Limit of e^{-z} as z Approaches 1+3i

  • Thread starter Mechdude
  • Start date
  • Tags
    Limit
In summary, substituting does not carry the full 3 marks on the homework. You will need to use the delta-epsilon definition to prove that the limit is really L.
  • #1
Mechdude
117
1

Homework Statement



whats the [tex] \lim_{z\rightarrow 1+3i} e^{-z} [/tex]



Homework Equations




do i simply substitute?

The Attempt at a Solution

I am out the door on this
 
Physics news on Phys.org
  • #2
It depends on how you are asked to do it.
Do you need to prove this is the limit using delta-epsilon definition?

Anyway the first step would be to evaluate it intuitivley- as you said, substitute.

Then if you are asked to prove, you will need to use the definition of the limit with your result.
 
  • #3
elibj123 said:
It depends on how you are asked to do it.
Do you need to prove this is the limit using delta-epsilon definition?

Anyway the first step would be to evaluate it intuitivley- as you said, substitute.

Then if you are asked to prove, you will need to use the definition of the limit with your result.

well I am only asked to evaluate, but the question carries 3 marks , could substitution carry all those marks? How would the epislon delta proof proceed?
 
  • #4
Once you know:

[tex]L=e^{-1+3i}[/tex] (whatever number it is)

You will have to plug it into the definition

For each [tex]\epsilon[/tex]>0 there exists such [tex]\delta[/tex]>0 so for any z which satisfies [tex]|z+1-3i|<\delta[/tex] it's true that [tex]|e^{z}-e^{-1+3i}|<\epsilon[/tex].

let z=x+iy, so:

[tex]|e^{x+iy}-e^{-1+3i}|=\frac{1}{e}|e^{3i}||e^{(x+1)+i(y-3)}-1|=
\frac{1}{e}\sqrt{(e^{x+1}cos(y-3)-1)^{2}+(e^{x+1}sin(y-3))^{2}}=(*)[/tex]

I used the fact that [tex]|e^{i\theta}|=1[/tex] for any real theta.

Now y is a real number, therefore cos(y-3) and sin(y-3) are bounded by 1. So:

[tex](*) \leq \frac{1}{e}\sqrt{(e^{x+1}-1)^{2}+e^{2(x+1)}} \leq \frac{1}{e}\sqrt{2e^{2(x+1)}}=\frac{\sqrt{2}}{e}e^{x+1}=(*)[/tex]

Notice that:

[tex]|x+1| \leq \sqrt{(x+1)^{2}+(y-3)^{2}} = |z+1-3i| < \delta [/tex]

Therefore since delta is positive:
[tex]x+1< \delta[/tex]

The exponent is monotonously increasing, so

[tex](*) < \frac{\sqrt{2}}{e}e^{\delta}[/tex]

Now the last expression is as small as we want it to be (except for being zero, but the limit definition doesn't require it), so for every choise of epsilon, we can choose such small delta, for which

[tex]|e^{z}-e^{-1+3i}|< \frac{\sqrt{2}}{e}e^{\delta}< \epsilon[/tex]

Therefore the limit is really L.
 

What is the definition of a limit in mathematics?

In mathematics, the limit of a function is the value that the function approaches as the input variable gets closer and closer to a specific value or approaches infinity. It is denoted by the symbol "lim".

How do you calculate the limit of a complex function?

To calculate the limit of a complex function, we treat the complex variable as two separate variables, one representing the real part and the other representing the imaginary part. Then, we calculate the limit of each part separately and combine them to get the final limit.

What is the difference between approaching a limit on the real axis and the imaginary axis?

When approaching a limit on the real axis, we are only considering the values of the function as the input variable approaches a real number. However, when approaching a limit on the imaginary axis, we are also considering the values of the function as the input variable approaches a purely imaginary number (such as 3i).

Can you take the limit of a complex function at a point where the function is not defined?

Yes, the limit of a complex function can still be calculated at a point where the function is not defined. This is because the limit is based on the behavior of the function as the input variable gets closer to the specified point, not necessarily the value of the function at that point.

What is the limit of e^{-z} as z approaches 1+3i?

The limit of e^{-z} as z approaches 1+3i is equal to e^{-1-3i} = e^{-1}e^{-3i} = e^{-1}(cos(-3) + isin(-3)) = e^{-1}(cos3 - isin3).

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
916
  • Calculus and Beyond Homework Help
Replies
6
Views
398
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
531
  • Calculus and Beyond Homework Help
Replies
10
Views
780
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
20
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
647
Back
Top