nate9228 said: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[itex]\pi[/itex]+2[itex]\pi[/itex])/5)-e^-((4i[itex]\pi[/itex]+2[itex]\pi[/itex])/5). This I am not so sure how to solve.
nate9228 said: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 said: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.
nate9228 said: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.
The integral question for cos(2z) in complex analysis is a mathematical problem that involves finding the integral (or area under the curve) of the complex function cos(2z) over a specific interval.
To solve the integral question for cos(2z) in complex analysis, you can use techniques such as substitution, integration by parts, or the Cauchy integral theorem. The specific method used will depend on the complexity of the function and the interval over which the integral is being evaluated.
The integral question for cos(2z) in complex analysis is important because it allows us to calculate the area under a complex function, which can have practical applications in physics, engineering, and other fields. It also helps us understand the behavior of complex functions and their relationship to real functions.
Yes, the integral question for cos(2z) in complex analysis can be solved using software such as Mathematica, MATLAB, or Maple. These programs have built-in functions and algorithms for solving complex integrals and can provide accurate and efficient solutions.
Some common mistakes to avoid when solving the integral question for cos(2z) in complex analysis include forgetting to account for the complex component of the function, using incorrect substitution or integration techniques, and making errors in the calculations. It is important to carefully check the steps and solutions to ensure accuracy.