Solve e^{\frac{-i \pi}{4}} = ?

  • Thread starter Bacat
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In summary, the conversation discusses the identities e^{i\pi}+1=0 and e^{i \pi} = e^{-i \pi}, and the difficulty in understanding e^{\frac{-i \pi}{4}} and its representation in the complex plane. The concept of rounding and different representations of the same complex number are also mentioned. The conversation concludes with the clarification that e^{\frac{-i \pi}{4}} does not equal 1.
  • #1
Bacat
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Not really homework, but part of a homework problem I am working on.

I know that [tex]e^{i\pi}+1=0[/tex] (Euler's Identity)

And also that [tex]e^{i \pi} = e^{-i \pi}[/tex]

But I'm having trouble understanding [tex]e^{\frac{-i \pi}{4}}[/tex]

In the complex plane this is a clockwise rotation around the origin of [tex]\frac{\pi}{2}[/tex] radians. But I think it should reduce to some real constant which I am having trouble finding.

In Mathematica, I get two different answers...

[tex]N[e^{\frac{-i \pi}{4}}] = 0.707107 - 0.707107 i[/tex]

This implies that [tex]e^{\frac{-i \pi}{4}} = \frac{1}{\sqrt{2}}(1-i)[/tex] which seems wrong to me.

The other answer given is:

Simplify[tex][e^{\frac{-i \pi}{4}}] = -(-1)^{\frac{3}{4}}[/tex]

But this reduces to 1, which I believe is probably the correct answer.

Is the first result just spurious rounding?

Can I just write the following identity as true?

[tex]e^{\frac{-i \pi}{4}} = 1[/tex]
 
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  • #2
e^(-i*pi/4) isn't 1 by a long shot. e^(i*x) is cos(x)+i*sin(x). e^(-i*pi/4)=cos(-pi/4)+i*sin(-pi/4). Both of your Mathematica answers are correct. In fact they are the same (after rounding). And (-1)^(3/4) is NOT 1. Or -1 either.
 
  • #3
Eh the first answer looks correct to me, since .707 is around sqrt(2)/2. You do know the Euler's formula, which is where those identities are usually derived from right?
 
  • #4
Bacat said:
Not really homework, but part of a homework problem I am working on.

I know that [tex]e^{i\pi}+1=0[/tex] (Euler's Identity)

And also that [tex]e^{i \pi} = e^{-i \pi}[/tex]

But I'm having trouble understanding [tex]e^{\frac{-i \pi}{4}}[/tex]

In the complex plane this is a clockwise rotation around the origin of [tex]\frac{\pi}{2}[/tex] radians.
No, the rotation is by [tex]\frac{\pi}{4}[/tex] radians. Was the 2 in the denominator a typo?
Bacat said:
But I think it should reduce to some real constant which I am having trouble finding.
Why would you think that? Just because [tex]e^{-i \pi}[/tex] is a real constant, doesn't mean that the other one is also a real constant.
Bacat said:
In Mathematica, I get two different answers...

[tex]N[e^{\frac{-i \pi}{4}}] = 0.707107 - 0.707107 i[/tex]
This implies that [tex]e^{\frac{-i \pi}{4}} = \frac{1}{\sqrt{2}}(1-i)[/tex] which seems wrong to me.
These are two representations of the same complex number. The first is an approximation and the second is exact.
Bacat said:
The other answer given is:

Simplify[tex][e^{\frac{-i \pi}{4}}] = -(-1)^{\frac{3}{4}}[/tex]
I am not familiar with Mathematica, so I don't know the difference between the N command and the Simplify command.
Bacat said:
But this reduces to 1, which I believe is probably the correct answer.
No it doesn't, and 1 is not the correct answer. Let's look at -(-1)3/4 a little more closely. Before the final sign change, you have (-1) to the 3/4 power. That is the same as -1 cubed (still -1), which we take the 4th root of. This is not a real number, since there is no real number that when squared, and then squared again, yields a negative number. The final step is to change the sign of this (nonreal) number, which still doesn't give us 1, as you claimed.
Bacat said:
Is the first result just spurious rounding?

Can I just write the following identity as true?

[tex]e^{\frac{-i \pi}{4}} = 1[/tex]
Absolutely not.
 
  • #5
Thanks Mark44 and all. This makes a lot more sense to me now.
 

What is e^{\frac{-i \pi}{4}}?

e^{\frac{-i \pi}{4}} is a mathematical expression that represents a complex number in the form of e raised to the power of a fraction with a complex number in the exponent.

What does the notation "i" in e^{\frac{-i \pi}{4}} stand for?

"i" in this notation represents the imaginary unit, which is equal to the square root of -1. It is commonly used in complex numbers to denote the presence of an imaginary component.

How do you solve e^{\frac{-i \pi}{4}}?

To solve e^{\frac{-i \pi}{4}}, we can use the polar form of complex numbers, where the magnitude is given by e, and the argument is given by -\frac{\pi}{4}. This can be expressed as e^{-\frac{\pi}{4}i}. Therefore, the solution is e^{-\frac{\pi}{4}i} = \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i.

What is the significance of e^{\frac{-i \pi}{4}}?

e^{\frac{-i \pi}{4}} is a commonly used complex number in many mathematical and scientific applications. It is particularly important in the fields of physics and engineering, where it is used to represent oscillatory phenomena such as electromagnetic waves and quantum mechanical systems. It also has significance in the study of complex analysis and trigonometry.

Can e^{\frac{-i \pi}{4}} be simplified further?

No, e^{\frac{-i \pi}{4}} is already in its simplest form as a complex number. However, it can be expressed in different forms such as the rectangular form (a + bi) or the exponential form (re^{i\theta}).

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