By using the definition of limits, we need to prove that, lim (6z = - = 2 + 6i z-1+i
By using the definition of limits, we need to prove that, lim (6z = - = 2 + 6i z-1+i
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Complex Analysis - Limits
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Transcribed Image Text:Math / Advanced Math / Q&A Library / Use the definition of limits to prove (6z-4)=2+6i as the limit of z approache...
Use the definition of limits to prove (6z-4)=2+6i as the limit of z approaches 1+i
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Step 1
By using the definition of limits, we need to prove that,
lim
6z
= 2 + 6i
z-1+i
Step 2
Assume that,
e > 0 is a small constant
Take another constant as
8 = £
6.
Assume that,
0 < |z – (1+ i)| < 8
Multiply with 6 on both sides
0 < [6z – 6 – 6i| < 68
0 < |6z – 4 – (2+ 6i)| < ɛ
I dont get this part, please explain
further or show the exact solution
-
As the limit tends to a small cons tant
By the deinition of limit, we get,
lim (6z – 4) = 2 + 6i
z-1+i
Hence proved.
![Example 1. Use the definition of limits to show that lim,-1-i[x+i(2x +y)I = 1+i.
Proof: Let ɛ> 0. Choose 8 =
Clearly 8 > 0.
3.
Assume
0 < ]z- (1-i) < 8,
where z x+ iy and
X,y E R.
Then
|x-1+i(y+ 1)| < 8.
Since
|x- 1 < |x-1+i(y+1)|< 8.
Thus
|[x+i(2x + y)] - (1+ i)| = |x – 1+i(y+ 1) +2i(x – 1)|
<|x-1+i(y+ 1)|+ |2i(x- 1)|
<&+12il|x- 1|
<8+ 28 = 38 = ɛ.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc26c26be-76f2-4a8a-a9a6-1bbb947bcf52%2F210eb4cd-7528-41d8-827e-0775495dab01%2F0q6wzql_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 1. Use the definition of limits to show that lim,-1-i[x+i(2x +y)I = 1+i.
Proof: Let ɛ> 0. Choose 8 =
Clearly 8 > 0.
3.
Assume
0 < ]z- (1-i) < 8,
where z x+ iy and
X,y E R.
Then
|x-1+i(y+ 1)| < 8.
Since
|x- 1 < |x-1+i(y+1)|< 8.
Thus
|[x+i(2x + y)] - (1+ i)| = |x – 1+i(y+ 1) +2i(x – 1)|
<|x-1+i(y+ 1)|+ |2i(x- 1)|
<&+12il|x- 1|
<8+ 28 = 38 = ɛ.
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