# Can someone explain this to me (First year algebra)

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#1
Having a bit of trouble with this idea since I missed the lecture.

If z is a complex number,

abs(e^-iz) is equal to...

I have expressed z as x+yi, and expanded that in place of the z, but i'm stuck here. I think i'm missing a rule or something.
0
6 years ago
#2
(Original post by Economi)
Having a bit of trouble with this idea since I missed the lecture.

If z is a complex number,

abs(e^-iz) is equal to...

I have expressed z as x+yi, and expanded that in place of the z, but i'm stuck here. I think i'm missing a rule or something.
What is the abs value of e^{-ix}? You can use e^{-ix} = cos x - i sin x.
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#3
(Original post by steve44)
What is the abs value of e^{-ix}? You can use e^{-ix} = cos x - i sin x.
I'm not completely sure
0
6 years ago
#4
(Original post by Economi)
I'm not completely sure
Use abs(a+ib) = \sqrt{a^2 + b^2}.
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#5
(Original post by steve44)
Use abs(a+ib) = \sqrt{a^2 + b^2}.
I appreciate you trying to coach me along but I don't have a very solid knowledge of complex numbers, and their various forms (although I should).

I'm going to need more tips.
0
6 years ago
#6
O.K. A complex number is a + i b (a and b are real numbers) where a and b are basically the x and y coordinates of a point in the complex plane. The abs value of a + i b is basically its distance from the origin. That is why I wrote abs(a+ib) = \sqrt{a^2 + b^2}.

So if e^{-ix} = cos x - i sin x

then

abs (e^{-ix}) = \sqrt{cos^2 x + (- sin x)^2} = \sqrt{cos^2 x + sin^2 x} = 1.

(To derive e^{-ix} = cos x - i sin x is to do with Taylor expansions).
0
#7
(Original post by steve44)
O.K. A complex number is a + i b (a and b are real numbers) where a and b are basically the x and y coordinates of a point in the complex plane. The abs value of a + i b is basically its distance from the origin. That is why I wrote abs(a+ib) = \sqrt{a^2 + b^2}.

So if e^{-ix} = cos x - i sin x

then

abs (e^{-ix}) = \sqrt{cos^2 x + (- sin x)^2} = \sqrt{cos^2 x + sin^2 x} = 1.

(To derive e^{-ix} = cos x - i sin x is to do with Taylor expansions).
I've gotten this far i just can't see how it gives me the answer to the question.
0
6 years ago
#8
O.K. So all you need to do is take e^{-iz} = e^{y - ix} = e^y e^{-ix} and use that abs(e^{-iz}) = abs(e^y e^{-ix})= e^y abs (e^{-ix}) from \sqrt{a^2 +b^2}.
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#9
(Original post by steve44)
O.K. So all you need to do is take e^{-iz} = e^{y - ix} = e^y e^{-ix} and use that abs(e^{-iz}) = abs (e^y e^{-ix}) from \sqrt{a^2 +b^2}.
How didn't i see that. I think I love you. Thanks

<3
0
6 years ago
#10
(Original post by Economi)
How didn't i see that. I think I love you. Thanks

<3
Steady on. A thumbs would do.
3
6 years ago
#11
(Original post by Economi)
Having a bit of trouble with this idea since I missed the lecture.

If z is a complex number,

abs(e^-iz) is equal to...

I have expressed z as x+yi, and expanded that in place of the z, but i'm stuck here. I think i'm missing a rule or something.
So started in the right place.

We now have: (e^-ix+y)
=(e^-ix)(e^y)

This is of the form z=Re^i(theta) so....
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#12
(Original post by natninja)
So started in the right place.

We now have: (e^-ix+y)
=(e^-ix)(e^y)

This is of the form z=Re^i(theta) so....
It's ok, i've gotten it now thanks to steve
0
6 years ago
#13
(Original post by natninja)
So started in the right place.

We now have: (e^-ix+y)
=(e^-ix)(e^y)

This is of the form z=Re^i(theta) so....
So it is of the form z = A e^{-ix}, where e^{-ix} represents a vector of unit length, therefore z = A e^{-ix} is a vector of length A(=e^y).
0
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