Question

If C is the triangle with vertices z=0, z=-4, and z=3i, what is the length of C? Please help!

Answer 1

bfore clearing u r problem do u know what type of math is this??

Answer 2

is the picture, I'm pretty sure

This is a complex calculus problem, and a small part of a larger question I am working on

Answer 3

(it didn't come out very well, but thats a triangle from -4 to 3i to zero)

Answer 4

u'll find difficulties as some what it is unrognized

Answer 5

What do you mean by length of triangle??

Answer 6

Sorry- Length of C, ie, the contour of the triangle

Answer 7

@feldy90 go to this link! - http://www.webmath.com/

Answer 8

sorry, I have never heard of contour of triangle. Can you give a better term for it??

Answer 9

The entire question is somewhat more complex than this... That is, show \[| \int\limits_{c}^{} e^z-zdz|\le60\] where C is the triangle with vertices z=0, z=-4, z=3i
And the method I am using is \[|\int\limits_{c}^{}e^z-zdz|\le \max(z \in C)|e^z-z|.\int\limits_{c}^{}|dz|\] And it is this final dz term which is the length of the contour. This is what my current question is about... Hope this clears anything up, and someone is still able to help?

Answer 10

@dumbcow you have helped me quite a lot with various types of calculus before... Any insight to offer on this one? :)

Answer 11

@Rohangrr thanks for the link, but I think my problem is slightly advanced for that site...

Answer 12

no short it up! @feldy90

Answer 13

hmm don't know a lot regarding complex analysis...let me see if i study up a bit or find something to help you

Answer 14

@feldy90 , is this your question

\[\int\limits_{c}^{} \left| e ^{z} -z \right|dz \le60\]

Because the position of modulus in your question is not clear.

Answer 15

No, the modulus is indicated as being outside the integral, ie\[\left| \int\limits_{c}^{} e^z-zdz \right| \le 60\] Also, the z term is actually the conjugate of z (z-bar, if you like)

Answer 16

And \[\left| \int\limits_{c}^{} e^z-zdz \right|\le \left(\begin{matrix}\max \\ z \in C\end{matrix}\right)\left| e^z-z \right|*\int\limits_{c}^{}\left| dz \right|\] Since this is true, I must compute the right hand side of this expression, and show that it is less than 60...

Answer 17

@feldy90 , see this link
--> http://mathhelpforum.com/calculus/77258-inequality-absolute-value-complex-integral.html

This can help you I suppose.

Answer 18

Answer = -2 + 2i
interpreted Formula: (1+i)^3

Answer 19

@feldy90

Answer 20

@feldy90 , I got the thing you are looking for.

See here
--> http://www.math.ust.hk/~maykwok/courses/ma304/06_07/Complex_4.pdf

Answer 21

Remember, with complex variables, keep like terms involving i together....
1+i evaluates to 1+i
Raising 1+i to the exponent 3 is the same as multiplying 1+i by itself 3 times.

Multiply 1+i by 1+i

Multiplying 1+i by 1+i is a classic Algebra problem. Here, you are trying
to multiply two binomials together (two expressions that each contain two terms).
Your book might call this finding the "Product of Two Binomials".

To work this problem, we'll use the "F.O.I.L." method. F.O.I.L. stands
for First, Outer, Inner, Last.

First, we'll multiply the two First terms, the 1 and 1 together.


1

1 × 1 = 1

Second, we'll multiply the two Outer terms, the 1 and i together.

Multiply 1 and i

Multiply 1 and i


The i just gets copied along.

i

1 × i = i

Third, we'll multiply the two Inner terms, the i and 1 together.

Multiply i and 1

Multiply i and 1


The i just gets copied along.

The answer is i

i

i × 1 = i

i combines with i to give 2i

Lastly, we'll multiply the two Last terms, the i and i together.

Multiply i and i

Multiply the i and i

Multiply i and i


Combine the i and i by adding the exponents, and keeping the i, to get

The answer is



i × i =



Multiply by 1+i

Multiply by 1+i

we multiply by each term in 1+i term by term.

This is the distributive property of multiplication.


1

1 × 1 = 1

Multiply 1 and i

Multiply 1 and i


The i just gets copied along.

i

1 × i = i

Multiply 2i and 1

Multiply i and 1


The i just gets copied along.

The answer is i

i

2i × 1 = 2i

i combines with 2i to give 3i

Multiply 2i and i

Multiply the i and i

Multiply i and i


Combine the i and i by adding the exponents, and keeping the i, to get

The answer is



2i × i =

Multiply and 1

Multiply and 1


The just gets copied along.

The answer is



× 1 =

combines with to give

Multiply and i

Multiply the and i

Multiply and i


Combine the and i by adding the exponents, and keeping the i, to get

The answer is



× i =



(1+i)^3 evaluates to


The final answer (almost!) is

Now, let's simplify the i's to get our final answer:
The i in 3i cannot be simplified, so just leave it as is.

The i2 in can be simplified to -1.

So, the becomes -3 .

1 combines with -3 to give -2

The i3 in can be simplified to -i.

So, the becomes -i .

3i combines with -i to give 2i


The final answer is
-2+2i

Answer 22

@Rohangrr This is taken from the webmath site you recommended, yes? It's not the addition or subtraction or multiplication of complex numbers which is my problem. I am wondering which vertices to add in the case of the triangle above to find my dz term To me, all I should be doing is summing the vertices, ie: 4 + 3i + sqrt(4^2 + 3i^2) (from pythagoras). This does not give me a reasonable answer, thus my asking for help from openstudy. Thank you, but I think we misunderstand each other...