Feb 16, 2007 #1 M mtts Junior Member level 3 Joined Nov 26, 2006 Messages 31 Helped 15 Reputation 30 Reaction score 13 Trophy points 1,288 Activity points 1,406 Hello has all I seek the solution of this equation. cos(x+iy)= a+ib with a and b are given (and x,y,a,b are real) . - how can one find x and y. - has which condition I have the place to make : x+iy= ±acos(a+ib ) +2πm thank you well
Hello has all I seek the solution of this equation. cos(x+iy)= a+ib with a and b are given (and x,y,a,b are real) . - how can one find x and y. - has which condition I have the place to make : x+iy= ±acos(a+ib ) +2πm thank you well
Feb 16, 2007 #2 J jayc Member level 3 Joined Feb 14, 2006 Messages 64 Helped 11 Reputation 22 Reaction score 0 Trophy points 1,286 Activity points 2,007 Re: cos complex The most straightforward way is to make use of Euler's Formula. See Euler's formula - Wikipedia, the free encyclopedia Basically, we can decompose a cos into a sum of complex exponentials: \[\cos(w) = \frac{e^{iw}+ e^{-iw}}{2}\] So just plug in "x + iy" for "w" and find the real and imaginary parts to equate with "a" and "b", respectively.
Re: cos complex The most straightforward way is to make use of Euler's Formula. See Euler's formula - Wikipedia, the free encyclopedia Basically, we can decompose a cos into a sum of complex exponentials: \[\cos(w) = \frac{e^{iw}+ e^{-iw}}{2}\] So just plug in "x + iy" for "w" and find the real and imaginary parts to equate with "a" and "b", respectively.
Mar 8, 2007 #3 H hamidrezakarami Full Member level 3 Joined Feb 26, 2007 Messages 180 Helped 45 Reputation 88 Reaction score 41 Trophy points 1,308 Activity points 2,162 Re: cos complex there is another method to solve it ... cos(x + jy) = cos(x).cos(jy) - sin(x).sin(jy) = cos(x).cosh - j.sin(x).sin = a +j.b => cos(x).cosh = a, and sin(x).sin = -b you can solve them to obtain final result.
Re: cos complex there is another method to solve it ... cos(x + jy) = cos(x).cos(jy) - sin(x).sin(jy) = cos(x).cosh - j.sin(x).sin = a +j.b => cos(x).cosh = a, and sin(x).sin = -b you can solve them to obtain final result.