What is the Solution for Finding a Real Scalar in a Complex Number Equation?

[DmytriE]

Homework Statement

u = -1 + j$\sqrt{3}$
v = $\sqrt{3}$ - j

Let a be a real scaling factor. Determine the value(s) of a such that

|u-$a/v$| = 2$\sqrt{2}$

Homework Equations

The equation above is the only relevant equation.

The Attempt at a Solution

I have converted the cartesian equation into polar in the hopes that it would be made easier but apparently not. I have gotten the following answer -8$\sqrt{3}$ + 12j. However, this does not work and is not a real scalar either...

This problem should be able to be done by hand.

 

[NascentOxygen]

You haven't shown any working, so it's difficult to point out where you went wrong.

Can you evaluate u-a/v and express it as a complex number?

 

[DmytriE]

u = -1 + j$\sqrt{3}$
v = $\sqrt{3}$ - j
|u - a/v| = 2$\sqrt{2}$

Here is what I have done step by step to get my answer.

$\sqrt{u - a/v}$ = 2$\sqrt{2}$

u - a/v = 8
v(u - 8) = a

Substitute in v and u and begin performing basic algebra.

($\sqrt{3}$ - j) * (-1 + j$\sqrt{3}$ - 8) = a
($\sqrt{3}$ - j) * (-9 + j$\sqrt{3}$) = a

Then FOIL the binomial
-9$\sqrt{3}$ + 3j + 9j + $\sqrt{3}$
a = -8$\sqrt{3}$ + 12j

Now, this is the complex number that I get but this is not a real scalar. How should I proceed or should I begin trying something else? This is part of the section where a calculator is not needed.

Food for thought:
2$\sqrt{2}$ can easily be expressed using trigonometric functions (sin($\frac{\pi}{4}$) and cos($\frac{\pi}{4}$))but I don't know how this can play a part.

 

[gneill]

u = -1 + j$\sqrt{3}$
v = $\sqrt{3}$ - j

Here is what I have done step by step to get my answer.

$\sqrt{u - a/v}$ = 2$\sqrt{2}$

Ah, but that's not the magnitude of the expression. For a complex number z = x + y*j, the magnitude is given by

$$|z| = \sqrt{x^2 + y^2} $$

Start by expanding the expression u - a/v and collect into its real and imaginary parts (assume that a is a real number). Then apply the definition of the magnitude to the result. Note that you can clear the square root by taking the square on both sides...

 

[DmytriE]

Start by expanding the expression u - a/v and collect into its real and imaginary parts (assume that a is a real number). Then apply the definition of the magnitude to the result.

Yes! Thank you gneill! I was blinded by my continuous mistakes. The help that made it all clear.