Harder Complex Numbers Question

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I'd first evaluate (simplify) [MATH]\frac{5-i}{\exp\left(\frac{7i\pi}{4}\right)}[/MATH]. Then what you've done should tie in.
 
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?
 
Dr.Peterson said:
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?
Click to expand...
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 ?
 
jasmeetcolumbia98 said:
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 ?
Click to expand...

The first line is nonsense. The next line is what I expected. The third line is in the wrong quadrant. Your suggestion to use the conjugate is the right thing to do, once you get the signs right!
 
Oh would I take pi out to make it cos(3pi/4)+isin(3pi/4)
isin(3pi/4)=sqrt(2)/2
cos(3pi/4)=-sqrt(2)/2
sqrt(2)/2 (-1+i)
(5-i)/(sqrt(2)/2 (-1+i))
then multiply top and bottom by
sqrt(2)i/2 + sqrt(2)/2
for the denominator if I multiply it out by
sqrt(2)i/2 + sqrt(2)/2
then I get 1/2+1/2(i^2) which would cancel out since i^2=1
 
Your "taking pi out" actually multiplied the number by -1! Don't change the quadrant. Just take it directly: what are the cosine and sine of 7 pi/4?

Then something went wrong in your multiplication; you'd better show more steps of that, so that either you or I can recognize the error. The product can't possibly be zero, can it?

To make the work less error-prone, I would suggest pulling the radical outside of the fraction, so that when you multiply by a conjugate, it will just be (1+i).
 
what are the cosine and sine of 7 pi/4?
+sqrt(2)/2 and -sqrt(2)/2
(sqrt(2)/2 - sqrt(2)i/2)((sqrt(2)/2 + sqrt(2)i/2)
1/2-(1/2i^2)=1
so for the denominator I now have 1
for the numerator
(5-i)*((sqrt(2)/2 + sqrt(2)i/2)
=5sqrt(2)/2 + 5sqrt(2)I/2 - isqrt(2)/2 -(sqrt(2)i^2)/2
which simplifies to
(sqrt(2)i^2)/2 + 5sqrt(2)/2 + 4sqrt(2)i/2
 
Okay, 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.
 
Dr.Peterson said:
Okay, 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.
Click to expand...
Ohhh yes I missed that out i^2=-1
(-sqrt(2))/2 + 5sqrt(2)/2 + 4sqrt(2)i/2
= 4sqrt(2)/2 +4isqrt(2)/2
Real part of this is 4sqrt(2)/2
Re(w)^2 conjugate = 4sqrt(2)/2
 
Check your signs, as I hinted! (My guess was that you may have thought the -1 but didn't replace i^2 with it.)
 
Yeah I replaced the sign without getting rid of [MATH]i^2[/MATH][MATH]Re(w)^2[/MATH] conjugate = [MATH]6sqrt(2)/2[/MATH]=[MATH]3sqrt2[/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 have assiduously avoided contributing to this thread.
I really dislike the question because I do not think it contributors to learning.
That said here are some related facts related to the post.
If \(z\ne 0\) then \(\dfrac{1}{z}=\dfrac{\overline{z}}{|z|^2}\).
And \(\forall(\theta)[\left|R\exp(i\theta)\right|=|R|\)
Moreover \(\Re(z)=\dfrac{z+z^*}{2}\) Sorry for the change of notation to \(z^*\) for conjugate. The LaTeX is off here.
 
c
Dr.Peterson said:
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 ...
Click to expand...
[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]
 
jasmeetcolumbia98 said:
[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]
Click to expand...
(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!
 
Dr.Peterson said:
(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!
Click to expand...
[MATH]cos(pi/4)= sqrt(2)/2 [/MATH]which is 1/6 of the
[MATH]Re(w)^2 conjugate 6sqrt(2)/2 =3sqrt2[/MATH]not sure where to go with this
 
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