Understanding I^(-i) and How to Solve for It

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In summary, I^(-i) is an imaginary number raised to the power of a complex number, resulting in a real number. It can be solved using Euler's formula and is important in various fields. It can also be simplified using exponent rules.
  • #1
PotatoSalad
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Really struggling with this question on my homework, can anybody help out? According to google the answer is 4.81047738 but I need to know how to get there.

Thanks for your help,
Pete.
 
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  • #2
Can you rewrite i in polar form using Euler's equation?
 
  • #3
e^(i.pi/2)?

I feel I'm getting close with some work I've done on paper.
 
  • #4
You're done. Now put that in for i (the base).
 
  • #5
Excellent, I've got it.

Thanks for your help neutrino
 

FAQ: Understanding I^(-i) and How to Solve for It

What is I^(-i)?

I^(-i) is an imaginary number raised to the power of a complex number, which results in a real number. In this case, the complex number is -i, which is the imaginary unit multiplied by -1.

How is I^(-i) different from i^(-i)?

I^(-i) and i^(-i) are both imaginary numbers raised to the power of a complex number. However, i^(-i) is a pure imaginary number while I^(-i) is a real number.

How do I solve for I^(-i)?

To solve for I^(-i), you can use the Euler's formula, which states that e^(ix) = cos(x) + i*sin(x). In this case, x = -1 and I^(-i) = e^(-i) = cos(-1) + i*sin(-1) = 0.5403 + 0.8414i.

Why is it important to understand I^(-i)?

I^(-i) is a fundamental concept in complex numbers and is used in various fields such as physics, engineering, and mathematics. It helps in solving problems involving oscillations, rotations, and waves.

Can I^(-i) be simplified?

Yes, I^(-i) can be simplified using the exponent rules for complex numbers. For example, I^(-i) = e^(-i) = e^(2*pi*i*i) = e^(-2*pi) = 0.1353.

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