Imaginary Numbers: Solve i^(4/3) Equation

  • Thread starter DivGradCurl
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In summary, to write i^{4/3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}, you can use the identity e^{i\theta} = \cos\theta + i\sin\theta and express it in Cartesian form. The result can also be obtained by using the formula i^{\frac{4}{3}} = e^{i(\frac{\pi}{2} + 2k\pi)(\frac{4}{3})} and taking the principal value. This value can be written as -\frac{1}{2} - i\frac{\sqrt{3}}{2} or 1 in Cartesian coordinates.
  • #1
DivGradCurl
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Folks, I was just wondering why I can write:

[tex]i^{4/3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}[/tex]

Regards
 
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  • #2
Do you know the identity [tex]e^{i\theta} = \cos\theta + i\sin\theta[/tex] ?

[tex]i = e^{i\frac{\pi}{2}}[/tex]

Use that, and the basic rules for exponentiation to get the result. Then express it in Cartesian form.
 
  • #3
Sure,complex exponentiation is multivalued,meaning that "i" can be written in an infinite ways using various arguments of the exponential...But worry about that less and use the formula which (though incomplete,hence inaccurate) was already given...

Daniel.
 
  • #4
dextercioby said:
Sure,complex exponentiation is multivalued,meaning that "i" can be written in an infinite ways using various arguments of the exponential...But worry about that less and use the formula which (though incomplete,hence inaccurate) was already given...

Daniel.

True, I know exponentiation is multivalued, but since he just wanted the principal value, that's what I gave him.
 
  • #5
Sorry to be nit-picking,but nowhere in his post is that "he just wanted the principal value" noticeable...

Daniel.

P.S.It doesn't matter,it's good if the OP got the simple part,at least.
 
  • #6
OK...

[tex]i^{\frac{4}{3}} = e^{i(\frac{\pi}{2} + 2k\pi)(\frac{4}{3})}[/tex]

The possible values in Cartesian coords are what was given, [tex]-\frac{1}{2} -i\frac{\sqrt{3}}{2}[/tex] and 1.

I know this. The point is, he just wanted help to "see" the first answer.
 
  • #7
thiago_j said:
Folks, I was just wondering why I can write:

[tex]i^{4/3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}[/tex]

Regards

You can write if you add "or 1, or

[tex]-\frac{1}{2}-i\frac{\sqrt{3}}{2}[/tex]",

as

[tex]i^{4/3}=(i^4)^{1/3}=\sqrt[3]{1} [/tex].

1 has got n different n-th roots,

[tex]\sqrt [n] {1} = \cos(\frac{2\pi}{n}k) + i \sin (\frac{2\pi}{n}k) [/tex]
k=0, ...n-1.
 
  • #8
Thank you for all the help! I think I finally get what you're saying...

[tex]z=i\, ^{\frac{4}{3}}=\left( i^4 \right)^{1/3} = \sqrt[3]{1}[/tex]

[tex]h= 1 = \mbox{cis } 0[/tex]

[tex]z=h_w ^3 = \mbox{ cis } \left( \frac{0+2k\pi}{3} \right) = \cos \left( \frac{2k\pi}{3} \right) +\mbox{ } i\mbox{ } \sin \left( \frac{2k\pi}{3} \right) = -\frac{1}{2} +\mbox{ } i \mbox{ } \frac{\sqrt{3}}{2}[/tex]
 

FAQ: Imaginary Numbers: Solve i^(4/3) Equation

What are imaginary numbers?

Imaginary numbers are numbers that when squared, give a negative result. They are denoted by the letter "i" and are used to represent the square root of a negative number.

What is the value of i^(4/3)?

The value of i^(4/3) is -1/√3 or -0.57735. This can be simplified to -√3/3.

How do you solve an equation with imaginary numbers?

To solve an equation with imaginary numbers, you can use the properties of exponents and the rules of complex numbers. In this case, we can use the formula i^(4/3) = (i^(1/3))^4 to simplify the equation.

Can imaginary numbers be used in real-life applications?

Yes, imaginary numbers have many real-life applications in fields such as engineering, physics, and economics. They are used to represent quantities that involve both real and imaginary components, such as alternating current in electrical circuits.

How do imaginary numbers relate to the complex plane?

Imaginary numbers are represented on the complex plane as the y-axis, with real numbers on the x-axis. The complex plane is a useful tool for visualizing and understanding complex numbers and their properties.

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