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Complex number equation

  • Thread starter Bob Busby
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  • #1
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Homework Statement



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

There are two different answer sets for {a,b} so find both of them.

Homework Equations



e^(j*a) = cos(a) + j*sin(a)

The Attempt at a Solution



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?
 

Answers and Replies

  • #2
vela
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OK, so you've rewritten the lefthand side in the form re where

[tex]\begin{align*}
r &= 5\sqrt{9+b^2} \\
\phi &= a+\tan^{-1} (b/3)
\end{align*}[/tex]

Now rewrite the righthand side the same way.
 
  • #3
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The righthand side would be sqrt(-25^2 + 0)*e^(j*arctan(0/-25) which is just 25*e^0. I don't see how that helps.
 
  • #4
vela
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The righthand side would be sqrt(-25^2 + 0)*e^(j*arctan(0/-25) which is just 25*e^0.
That's not right: 25e0 = +25, not -25.
I don't see how that helps.
If w and z are complex numbers such that w=z, then |w|=|z|, etc.
 
  • #5
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Oh yeah. What was wrong with the way I converted it? How come the minus sign wasn't preserved? Edit: Rather, how am I supposed to preserve the minus sign. I see nothing wrong with my conversion method.

"If w and z are complex numbers such that w=z, then |w|=|z|, etc."

But if I convert back to rectangular coordinates to take the magnitude then I will have a (cos(a+ arctan(b/3))^2 and I don't see how to do anything with that, even if it equals the magnitude of the other complex number.
 
  • #6
vela
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What are the geometric interpretation of r and ϕ? Where does -25 lie on the complex plane?
 
  • #7
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Hint:

[tex]
5 e^{i a} (3 + i b) = -25
[/tex]

[tex]
(\cos{(a)} + i \sin{(a)})(3 + i b) = -5
[/tex]

[tex]
(3 \cos{(a)} - b \sin{(a)}) + i (b \cos{(a)}+ 3 \sin{(a)}) = -5
[/tex]

[tex]
\left(\begin{array}{cc}
3 & -b \\

b & 3
\end{array}\right) \cdot \left(\begin{array}{c}
\cos{(a)} \\

\sin{(a)}
\end{array}\right) = \left(\begin{array}{c}
-5 \\

0
\end{array}\right)
[/tex]

[tex]
\left(\begin{array}{c}
\cos{(a)} \\

\sin{(a)}
\end{array}\right) = \frac{1}{9 + b^{2}} \left(\begin{array}{cc}
3 & b \\

-b & 3
\end{array}\right) \cdot \left(\begin{array}{c}
-5 \\

0
\end{array}\right)
[/tex]

[tex]
\begin{array}{l}
\cos{(a)} = -\frac{15}{9 + b^{2}} \\

\sin{(a)} = \frac{5 b}{9 + b^{2}}
[/tex]

Then, use the trigonometric identity:

[tex]
\cos^{2}{(a)} + \sin^{2}{(a)} = 1
[/tex]

substituting the above expressions and, after simplification, you have a biquadratic equation with respect to b. Once you find the solutions, substitute back in the expressions for [itex]\cos{(a)}[/itex] and [itex]\sin{(a)}[/itex] and find the angle which gives those values for the sine and the cosine (of course, up to an addtive factor of [itex]2\pi n[/itex]).
 
  • #8
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Wait a second. When I get b and try to solve for a I get two different answers. Is there a simpler method to solve with, though, without linear algebra?

"What are the geometric interpretation of r and ϕ? Where does -25 lie on the complex plane?"

r is the magnitude and phi is the angle ccw above the real axis. Doesn't where -25 lie depend on phi? Why wasn't the negative sign preserved in my conversion?
 
Last edited:
  • #9
vela
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To be a bit more precise, ϕ is measured from the positive real axis. This is just polar coordinates. The sign was wrong on your previous answer was because your ϕ was wrong. The point z=-25 lies on the negative real axis a distance 25 away from the origin, so r=25. What should ϕ equal?
 

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