# Complex number equation

## 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?

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vela
Staff Emeritus
Homework Helper
OK, so you've rewritten the lefthand side in the form re where

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

Now rewrite the righthand side the same way.

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
Staff Emeritus
Homework Helper
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.

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

Hint:

$$5 e^{i a} (3 + i b) = -25$$

$$(\cos{(a)} + i \sin{(a)})(3 + i b) = -5$$

$$(3 \cos{(a)} - b \sin{(a)}) + i (b \cos{(a)}+ 3 \sin{(a)}) = -5$$

$$\left(\begin{array}{cc} 3 & -b \\ b & 3 \end{array}\right) \cdot \left(\begin{array}{c} \cos{(a)} \\ \sin{(a)} \end{array}\right) = \left(\begin{array}{c} -5 \\ 0 \end{array}\right)$$

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

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

Then, use the trigonometric identity:

$$\cos^{2}{(a)} + \sin^{2}{(a)} = 1$$

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 $2\pi n$).

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:
vela
Staff Emeritus