Harder Complex Numbers Question

[Dr.Peterson]

[MATH]cos(pi/4)= sqrt(2)/2 [/MATH]which is 1/6 of the
[MATH]Re(w)^2 conjugate 6sqrt(2)/2 =3sqrt2[/MATH]

Keep going ...

and then check your answer.

 

[jasmeetcolumbia98]

Keep going ...

and then check your answer.

Would you do[MATH] sqrt(2)/2 - 3sqrt(2) = -5sqrt(2)/2 [/MATH]

 

[Dr.Peterson]

I'm not sure of some of the details of what you're thinking, but you're nearly there, so I'll show you where I am at this point, taking a slightly different route.

We want to find [MATH]|w|[/MATH] such that [MATH]Re\left(\frac{5-i}{\exp\left(\frac{7i\pi}{4}\right)} - \bar{w}^2\right) = 0[/MATH].

We've simplified the fraction, so this becomes [MATH]Re((3\sqrt{2}+2i\sqrt{2}) - \bar{w}^2) = 0[/MATH].

Now, since [MATH]Arg(w) = \frac{\pi}{8}[/MATH], we can write [MATH]w = R e^{\pi/8}[/MATH]. The goal is to find R, right?

You've found the real part of [MATH]\bar{w}^2[/MATH], if R = 1, to be [MATH]\frac{\sqrt{2}}{2}[/MATH]; so leaving my R in there, [MATH]Re(\bar{w}^2)= R^2\frac{\sqrt{2}}{2}[/MATH].

This turns our equation into [MATH]3\sqrt{2} - R^2\frac{\sqrt{2}}{2} = 0[/MATH].

Now solve for R.

(I have to agree with pka that this is more trouble than it's worth pedagogically; but at least it's giving you a chance to make all sort of little mistakes, which is a good experience if you learn from them ...)

 

[jasmeetcolumbia98]

Ohhhh I understand
3sqrt(2)=R^2 (sqrt(2)/2)
Ah so R^2 =6
R = sqrt(6)

 

[Dr.Peterson]

Yes, you've got it.

And to check (which is wise after all this), you can write out exactly what w ends up as, and evaluate the expression to see that it is, indeed, purely imaginary.

 

[jasmeetcolumbia98]

Yes, you've got it.

And to check (which is wise after all this), you can write out exactly what w ends up as, and evaluate the expression to see that it is, indeed, purely imaginary.

and R is essentially the modulus of w

 

[Dr.Peterson]

Yes, that is probably one of the key ideas in this exercise. (If you allow R to be negative, of course, then |w| = |R|.)