Solving complex exponentials

  • Thread starter Bob Busby
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I posted this problem here because I would like to know a reliable method for solving such a thing.

(5e^(j*a))(3 + j*b) = -25 Find real numbers a and b satisfying the preceding equation.

I converted it to get 5*sqrt(9 + b^2)*e^(j*a + j * arctan(b/3)) = -25. I don't really see where to go from here. If I seperate the real parts I will just get a cos(a + arctan(b/3)) which doesn't help me even if I equate real and imaginary parts. What do I do?
 

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  • #2
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I posted this problem here because I would like to know a reliable method for solving such a thing.

(5e^(j*a))(3 + j*b) = -25 Find real numbers a and b satisfying the preceding equation.

I converted it to get 5*sqrt(9 + b^2)*e^(j*a + j * arctan(b/3)) = -25. I don't really see where to go from here. If I seperate the real parts I will just get a cos(a + arctan(b/3)) which doesn't help me even if I equate real and imaginary parts. What do I do?
You try

[tex] e^{ja}= cos (a) + j sin(a) \hbox { and }\; 3+jb = \sqrt { 3^2 + b^2 } \; e^{j[tan^{-1} (\frac b 3)]} \hbox {?}[/tex]
 
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  • #3
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You try

[tex] e^{ja}= cos (a) + j sin(a) \hbox { and }\; 3+jb = \sqrt { 3^2 + b^2 } \; e^{j[tan^{-1} (\frac b 3)]} \hbox {?}[/tex]
Yes. that's what I tried.
 
  • #4
Redbelly98
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5*sqrt(9 + b^2)*e^(j*a + j * arctan(b/3)) = -25. I don't really see where to go from here.
How about rewriting that equation with -25 in the form c*e^(j*θ), where c is a positive real number?
 

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