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

[Mechdude]

Homework Statement

whats the $$\lim_{z\rightarrow 1+3i} e^{-z}$$

Homework Equations

do i simply substitute?

The Attempt at a Solution

I am out the door on this

 

[elibj123]

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.

 

[Mechdude]

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?

 

[elibj123]

Once you know:

$$L=e^{-1+3i}$$ (whatever number it is)

You will have to plug it into the definition

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

let z=x+iy, so:

$$|e^{x+iy}-e^{-1+3i}|=\frac{1}{e}|e^{3i}||e^{(x+1)+i(y-3)}-1|=<br /> \frac{1}{e}\sqrt{(e^{x+1}cos(y-3)-1)^{2}+(e^{x+1}sin(y-3))^{2}}=(*)$$

I used the fact that $$|e^{i\theta}|=1$$ for any real theta.

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

$$(*) \leq \frac{1}{e}\sqrt{(e^{x+1}-1)^{2}+e^{2(x+1)}} \leq \frac{1}{e}\sqrt{2e^{2(x+1)}}=\frac{\sqrt{2}}{e}e^{x+1}=(*)$$

Notice that:

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

Therefore since delta is positive:
$$x+1< \delta$$

The exponent is monotonously increasing, so

$$(*) < \frac{\sqrt{2}}{e}e^{\delta}$$

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

$$|e^{z}-e^{-1+3i}|< \frac{\sqrt{2}}{e}e^{\delta}< \epsilon$$

Therefore the limit is really L.