Integral question for cos(2z) in complex analysis

[nate9228]

Homework Statement

Evaluate ∫cos(2z)dz from pi/2 to pi/2+i

Homework Equations

The Attempt at a Solution

I know the cos function is entire, thus independent of path and I just need to evaluate the end points. I also know when you integrate you get (1/2)sin2z. My only question is for finishing the problem, how do I evaluate sin(pi/2+i). Its the i that is throwing me off.

 

[Dick]

sin(z)=(e^(iz)-e^(-iz))/(2i). You know how to do complex exponentials, right?

 

[nate9228]

Alright I figured it was that formula. So (1/2)sin(2z) becomes (1/4i)(e^(2iz)-e^(-2iz)). Yes I do know how to do complex exponentials , but not very well at a practical level yet. I did take the i out of the denominator of 2+i (using the conjugate) and then after putting into the formula I have (1/4i)(e^((4i\pi+2\pi)/5)-e^-((4i\pi+2\pi)/5). This I am not so sure how to solve.

 

[Dick]

Alright I figured it was that formula. So (1/2)sin(2z) becomes (1/4i)(e^(2iz)-e^(-2iz)). Yes I do know how to do complex exponentials , but not very well at a practical level yet. I did take the i out of the denominator of 2+i (using the conjugate) and then after putting into the formula I have (1/4i)(e^((4i\pi+2\pi)/5)-e^-((4i\pi+2\pi)/5). This I am not so sure how to solve.

Yes, that's the right formula, but I don't see how substituting pi/2+i for z gives what you've got. Where is the '/5' part coming from? And some other stuff doesn't look right. Show what you did. But once you've got it right, then just use Euler's formula, e^(a+bi)=e^a*(cos(b)+i*sin(b)).

 

[nate9228]

I did (pi/2+i)*(2-i)/(2-i)= (2pi-ipi)/5... specifically the 5 came from (2+i)(2-i)=(2^2)-i^2=5. I then took (2pi-ipi)/5 and plugged it in for z in the formula, which then has it being multiplied by 2i, and 2i*(2pi-ipi)/5= (4ipi+2pi)/5. Sorry for leaving all of that out.

 

[nate9228]

I did (pi/2+i)*(2-i)/(2-i)= (2pi-ipi)/5... specifically the 5 came from (2+i)(2-i)=(2^2)-i^2=5. I then took (2pi-ipi)/5 and plugged it in for z in the formula, which then has it being multiplied by 2i, and 2i*(2pi-ipi)/5= (4ipi+2pi)/5. Sorry for leaving all of that out.

 

[Dick]

I did (pi/2+i)*(2-i)/(2-i)= (2pi-ipi)/5... specifically the 5 came from (2+i)(2-i)=(2^2)-i^2=5. I then took (2pi-ipi)/5 and plugged it in for z in the formula, which then has it being multiplied by 2i, and 2i*(2pi-ipi)/5= (4ipi+2pi)/5. Sorry for leaving all of that out.

You are doing some seriously bad complex arithmetic there. (pi/2+i) is not equal to (2pi-i*pi)/5=2pi/5-i*pi/5. The real and imaginary parts are different. I'm not even sure what you think you are doing. Can you explain. Here's part of the right way.

If z=pi/2+i then exp(2iz)=exp(2*i*(pi/2+i))=exp(pi*i-2)=exp(-2)*(cos(pi)+i*sin(pi))=(-exp(2)). That's not so hard, yes? You are putting some kind of extra erroneous step in there.

 

[nate9228]

Sorry, I was at a loss earlier and did that based on what a friend said because I had no idea. It did make sense to me though so it was my own error. In regards to the arithmetic: I was under the impression when you have i in the denominator, to put the fraction in a+bi form, you multiply the fraction by the conjugate of the bottom. For example, if z= 2/(1+i) you multiply by (1-i)/(1-i) to get (2-2i)/2= 1-i. So I applied that same idea, albeit quite improperly apparently, to pi/(2+i). So basically I multiplied the top and bottom of pi/(2+i) by the conjugate of the denominator, i.e 2-i.

In regards to your answer: Once again, thank you. That's extremely simple. The only thing I'm not understanding at first glance is why does exp(2*i*(pi/2+i))= exp(pi*i-2)? I must say, I find it ironic I am able to prove some of the more challenging theorems, yet I apparently suck at simple complex arithmetic.

 

[Dick]

Sorry, I was at a loss earlier and did that based on what a friend said because I had no idea. It did make sense to me though so it was my own error. In regards to the arithmetic: I was under the impression when you have i in the denominator, to put the fraction in a+bi form, you multiply the fraction by the conjugate of the bottom. For example, if z= 2/(1+i) you multiply by (1-i)/(1-i) to get (2-2i)/2= 1-i. So I applied that same idea, albeit quite improperly apparently, to pi/(2+i). So basically I multiplied the top and bottom of pi/(2+i) by the conjugate of the denominator, i.e 2-i.

In regards to your answer: Once again, thank you. That's extremely simple. The only thing I'm not understanding at first glance is why does exp(2*i*(pi/2+i))= exp(pi*i-2)? I must say, I find it ironic I am able to prove some of the more challenging theorems, yet I apparently suck at simple complex arithmetic.

To put (a+bi)/(c+di) into rectangular form you multiply BOTH numerator and denominator by the conjugate of the denominator (c-di). But that's not the problem here. Just multiply it out. 2*i*(pi/2+i)=2*i*pi/2+2*i*i=i*pi-2. Maybe being good at the challenging problems is making you overcomplicate the simple ones.

 

[nate9228]

One last comment.. just because I realized the source of both our confusions as to what the other was doing. I was looking at the limit of integration wrong. Like utterly wrong lol. I thought it was pi divided by the quantity 2+i, when really, as you elucidated, it is (pi/2)+i. That is why I was so confused about the multiplication. I thought it was 2*i*(pi/(2+i))= (2ipi/(2+i)), when clearly its just the normal distributive property, 2*i*(pi/2)+2*i*i. Its amazing what some sleep and the taking another look will reveal.

 

[Dick]

One last comment.. just because I realized the source of both our confusions as to what the other was doing. I was looking at the limit of integration wrong. Like utterly wrong lol. I thought it was pi divided by the quantity 2+i, when really, as you elucidated, it is (pi/2)+i. That is why I was so confused about the multiplication. I thought it was 2*i*(pi/(2+i))= (2ipi/(2+i)), when clearly its just the normal distributive property, 2*i*(pi/2)+2*i*i. Its amazing what some sleep and the taking another look will reveal.

Ohh. So that's what you were doing. Seemed pretty mysterious, but I probably should have guessed.