What is the solution to the complex number equation (5e^(j*a))(3 + j*b) = -25?

[Bob Busby]

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 separate 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?

 

[vela]

OK, so you've rewritten the lefthand side in the form re^jϕ where

\begin{align*}<br /> r &amp;= 5\sqrt{9+b^2} \\<br /> \phi &amp;= a+\tan^{-1} (b/3)<br /> \end{align*}

Now rewrite the righthand side the same way.

 

[Bob Busby]

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.

 

[vela]

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.

 

[Bob Busby]

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.

 

[vela]

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

 

[Dickfore]

Hint:

<br /> 5 e^{i a} (3 + i b) = -25<br />

<br /> (\cos{(a)} + i \sin{(a)})(3 + i b) = -5<br />

<br /> (3 \cos{(a)} - b \sin{(a)}) + i (b \cos{(a)}+ 3 \sin{(a)}) = -5<br />

<br /> \left(\begin{array}{cc}<br /> 3 &amp; -b \\<br /> <br /> b &amp; 3<br /> \end{array}\right) \cdot \left(\begin{array}{c}<br /> \cos{(a)} \\<br /> <br /> \sin{(a)}<br /> \end{array}\right) = \left(\begin{array}{c}<br /> -5 \\<br /> <br /> 0<br /> \end{array}\right)<br />

<br /> \left(\begin{array}{c}<br /> \cos{(a)} \\<br /> <br /> \sin{(a)}<br /> \end{array}\right) = \frac{1}{9 + b^{2}} \left(\begin{array}{cc}<br /> 3 &amp; b \\<br /> <br /> -b &amp; 3<br /> \end{array}\right) \cdot \left(\begin{array}{c}<br /> -5 \\<br /> <br /> 0<br /> \end{array}\right)<br />

<br /> \begin{array}{l}<br /> \cos{(a)} = -\frac{15}{9 + b^{2}} \\<br /> <br /> \sin{(a)} = \frac{5 b}{9 + b^{2}}<br />

Then, use the trigonometric identity:

<br /> \cos^{2}{(a)} + \sin^{2}{(a)} = 1<br />

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 \cos{(a)} and \sin{(a)} and find the angle which gives those values for the sine and the cosine (of course, up to an addtive factor of 2\pi n).

 

[Bob Busby]

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?

 

[vela]

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?