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

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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...

The answer is e^Im(z).

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.

If z is a complex number,

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

The answer is e^Im(z).

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.

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#2

(Original post by

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...

The answer is e^Im(z).

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.

**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...

The answer is e^Im(z).

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.

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(Original post by

What is the abs value of e^{-ix}? You can use e^{-ix} = cos x - i sin x.

**steve44**)What is the abs value of e^{-ix}? You can use e^{-ix} = cos x - i sin x.

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#4

(Original post by

I'm not completely sure

**Economi**)I'm not completely sure

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(Original post by

Use abs(a+ib) = \sqrt{a^2 + b^2}.

**steve44**)Use abs(a+ib) = \sqrt{a^2 + b^2}.

I'm going to need more tips.

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#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).

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).

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(Original post by

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).

**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).

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#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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(Original post by

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}.

**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}.

<3

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#11

**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...

The answer is e^Im(z).

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.

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

=(e^-ix)(e^y)

This is of the form z=Re^i(theta) so....

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(Original post by

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....

**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....

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#13

**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....

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