jasmeetcolumbia98
New member
- Joined
- May 15, 2020
- Messages
- 42
it's conjugate would be x-iy
Arg(w)=arctan(y/x)
tan(-pi/8) = 1-sqrt(2)
1-sqrt(2)=(y/x)
I don't really know where to go or if this is right so far
e^(i) * e^(7pi/4)You can do much better than -(-1)^(3/4)! That, in fact, has four values.
What are the angle and magnitude of [MATH]e^\frac{7i\pi}{4}[/MATH]? What is that in a+ib form?
e^(i) * e^(7pi/4)
cos(7pi/4)+isin(7pi/4)
sqrt(2)/2 (1+i)
(5-i)/(sqrt(2)/2 (1+i))
would I multiply top and bottom by
(sqrt(2)/2) - (sqrt(2)/2)i ?
Ohhh yes I missed that out i^2=-1Okay, so you're saying the original denominator is (sqrt(2)/2 - sqrt(2)i/2), and then you're multiplying the numerator and denominator by ((sqrt(2)/2 + sqrt(2)i/2) . Correct.
You didn't quite finish the new numerator; what is [MATH]i^2[/MATH]? (There's also either a sign error or a typo.)
Now looking ahead to the next step, you'll want the real part of the whole expression to be zero, so you'll want to find [MATH]w[/MATH] so that the real part of [MATH]\bar{w}^2[/MATH] equals the real part of the fraction you're currently simplifying.
[MATH]Re(w)^2 = -3sqrt2[/MATH]not sure how I'd go about this and wouldn't I need to use Arg(w)=-pi/8 at some pointSounds good. Just finish up by finding w.
[MATH]cos(-pi/8)+isin(-pi/8)=cos(pi/8)-isin(pi/8) e^(-i*pi/8) [/MATH]then take conjugate which would be [MATH]e^(i*pi/8)[/MATH]squared would make it [MATH]e^(i*pi/4)[/MATH]maybe I could separate it to make [MATH](e^i)(e^(pi/4))[/MATH]making the real part [MATH]e^(pi/4)[/MATH]Yes. This is the part of the problem you were working on in the OP.
I would perhaps first find the complex number with magnitude 1 and argument -pi/8. I'd take its conjugate and square it. Then I'd look at the real part of this and think, what do I have to multiply this number by to get the required real part?
But check the sign in what you wrote here ...
(I've corrected some of your LaTeX.)[MATH]\cos(-\pi/8)+i\sin(-\pi/8)=\cos(\pi/8)-i\sin(\pi/8)= e^{-i\pi/8} [/MATH]then take conjugate which would be [MATH]e^{i\pi/8}[/MATH]squared would make it [MATH]e^{i\pi/4}[/MATH]maybe I could separate it to make [MATH](e^i)(e^{\pi/4})[/MATH]making the real part [MATH]e^{\pi/4}[/MATH]
[MATH]cos(pi/4)= sqrt(2)/2 [/MATH]which is 1/6 of the(I've corrected some of your LaTeX.)
You're making a mistake at the end that you made before. It is not true that [MATH]a^{mn} = a^ma^n[/MATH]! Rather, [MATH]a^{mn} = (a^m)^n[/MATH]. But that doesn't work here; and your conclusion about the real part is invalid.
The real part of [MATH]e^{i\pi/4}[/MATH] is [MATH]\cos{\pi/4}[/MATH]. Continue from there!