So there is another complex number question.

u = −1 + j√3 and v = √3 − j

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

|u −a/v | = 2√2

So this is what I am doing I am kind of stuck and wondering if I am on the right track.

Printable View

- Feb 10th 2013, 10:57 AMkmalik001Imaginary numbers again
So there is another complex number question.

u = −1 + j√3 and v = √3 − j

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

|u −a/v | = 2√2

So this is what I am doing I am kind of stuck and wondering if I am on the right track. - Feb 10th 2013, 10:58 AMkmalik001Re: Imaginary numbers again
- Feb 10th 2013, 11:15 AMMINOANMANRe: Imaginary numbers again
ok you are doing fine...continue....get the modulus of the complex number you found and then square it...and solve the quadratic equation that you will find...

finally you will get two solutions

a =2sqr(3)-2sqr(7) and a= 2sqr(7)-2sqr(3).

Good luck

Minoas - Feb 10th 2013, 11:35 AMkmalik001Re: Imaginary numbers again
That is kind of how I got stuck see while taking the modulas I have to mutliply this number by its conjugate. And that multiplication is messy unless I am doing it wrong.

This is what I did - Feb 10th 2013, 11:36 AMkmalik001Re: Imaginary numbers again
- Feb 10th 2013, 01:28 PMDevenoRe: Imaginary numbers again
|z| = √(Re(z)

^{2}+ Im(z)^{2})

in this case:

Re(z) = (1/4)(-4 - a√3)

Im(z) = (1/4)(4√3 - a)

squaring, we have:

Re(z)^{2}= (1/16)(16 + 8a√3 + 3a^{2})

Im(z)^{2}= (1/16)(48 - 8a√3 + a^{2})

adding these together, we get:

Re(z)^{2}+ Im(z)^{2}= (1/16)(64 + 4a^{2})

rather than deal with messy square roots, let's square both sides:

|z|^{2}= 2

(1/16)(64 + 4a^{2}) = 2

64 + 4a^{2}= 32

4a^{2}+ 32 = 0 <---this has no real solutions. - Feb 10th 2013, 02:18 PMkmalik001Re: Imaginary numbers again
Thank you I have sucessfully learned the modulus method now

just one thing I wanted to point out |z|^2 = 2√2 so then a comes out to be +- 4.

and the question said it wanted to be a real scaling factor so thats perfect. Thank you once again :)