Solving a Puzzling Equation: -e-x + e+x/2i

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SUMMARY

The discussion centers on the mathematical manipulation involving the imaginary unit \(i\) and the equation \(-e^{-x} + \frac{e^{x}}{2i}\). Participants clarify that \(\frac{1}{i}\) simplifies to \(-i\), which is crucial for understanding the equivalence of the two expressions. The transformation from the denominator to the numerator is explained through the relationship \(i^2 = -1\), leading to the conclusion that \(\frac{1}{i} = -i\). This foundational knowledge of complex numbers is essential for solving the equation correctly.

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  • Understanding of complex numbers, specifically the imaginary unit \(i\)
  • Familiarity with exponential functions and their properties
  • Basic algebraic manipulation skills
  • Knowledge of mathematical notation and simplification techniques
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  • Study the properties of imaginary numbers and their applications in complex analysis
  • Learn about exponential functions involving complex variables
  • Explore algebraic manipulation techniques for simplifying complex equations
  • Investigate the geometric interpretation of complex numbers on the complex plane
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Students studying mathematics, particularly those focusing on complex analysis, algebra, and anyone seeking to deepen their understanding of exponential functions involving imaginary numbers.

struggles
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Homework Statement


So this is a really simple problem and i know I'm missing something really obvious but i just can't spot it.

Screen Shot 2016-05-16 at 15.48.10.png


Homework Equations

The Attempt at a Solution


so in the second part above I get :
e-x - ex/2i. However I don't get the next bit as the 2i on the denominator is now factored into the top equation to give -ie-x + iex/2. I don't get how the 2 equations are equivalent and what they've done to get the i from the denominator to the numerator. If anyone could explain how this happened that'd be great!
 
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What is i?
 
Do you know what ##1/i## is?
 
axmls said:
Do you know what ##1/i## is?
i-1?
 
struggles said:
i-1?
First answer my question and then use it to answer struggles'. ##i^{-1}## is simply another form of ##\frac{1}{i}## which doesn't really help.
 
Suppose you didn't know what ##i## is
but it seems that (from your second equals sign) ##\frac {1}{i}= -i##... what must ##i## satisfy?
 
robphy said:
Suppose you didn't know what ##i## is
but it seems that (from your second equals sign) ##\frac {1}{i}= -i##... what must ##i## satisfy?
How is it defined? What stands ##i## for? There must have been some explanation for it.
Remark: Sorry, for confusing your names, struggles and axmls!
 
Though it isn't obvious that this is the way to find this, here's a hint:
$$\frac{1}{i} = \frac{1}{i} \frac{i}{i}$$
 
so i = √-1 so i2 = -1. so 1/i must equal 1/(-1)1/2 = (-1)-1/2? is that what you were getting at fresh-_42?
So axmls 1/i = i/i^2 = -i! Thanks both of you!
 

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