Understanding Exponential Complex Numbers

AI Thread Summary
The discussion focuses on understanding the expression e^{iθ} and its application to specific cases like e^{4i}. The correct evaluation of e^{4i} is clarified as cos(4) + i sin(4), where the argument for sine and cosine is determined by the coefficient of i. Participants also explore the relationship between different forms of complex exponentials, noting that e^{-ibx} can be expressed using trigonometric identities. The conversation emphasizes that both forms are equivalent through appropriate application of trigonometric identities. Overall, the thread enhances comprehension of exponential complex numbers and their evaluations.
adartsesirhc
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Homework Statement


I've never understood e^{i\theta} very well. I know that e^{i\theta} = cos \theta + i sin \theta, but how about e^{4i}? Would this be cos 1 + 4i sin 1 or cos 4 + i sin 4? What's the general rule for these kinds of numbers?


Homework Equations


e^{i\theta} = cos \theta + i sin \theta
e^{i\pi} + 1 = 0


The Attempt at a Solution


None, really. Just wondering how to evaluate numbers like above.
 
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I'm not sure where you got cos(1) + 4i*sin(1) from. For e^(4i), you would simply replace theta with 4, which gives the latter of the two expressions above. I assume we're dealing with radian measure so cos(4) + i*sin(4) can be evaluated using a calculator. On the complex plane, the point would be in the third quadrant since pi < 4 < (3/2)pi.
 
adartsesirhc said:

Homework Statement


I've never understood e^{i\theta} very well. I know that e^{i\theta} = cos \theta + i sin \theta, but how about e^{4i}? Would this be cos 1 + 4i sin 1 or cos 4 + i sin 4? What's the general rule for these kinds of numbers?
The "rule" is given in the formula you state: e^{i\theta}= cos(\theta)+ i sin(\theta). In e^{4i}[/itex], the number multiplying i is &quot;4&quot;: \theta= 4. e^{4i}= cos(4)+ i sin(4).<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <h2>Homework Equations</h2><br /> e^{i\theta} = cos \theta + i sin \theta<br /> e^{i\pi} + 1 = 0<br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> None, really. Just wondering how to evaluate numbers like above. </div> </div> </blockquote>
 
Hmm... so does this mean that whatever you multiply i by will be the argument of the sine and cosine? But my differential equations book has

e^{-ibx} = cos bx - i sin bx.

How do I know if this isn't

e^{-ibx} = e^{i(-bx)} = cos (-bx) + i sin (-bx)?

How do I tell when it's one and when it's the other?
 
OHHHH... never mind. =]

It just hit me: both are actually the same - just apply the appropriate trig identities.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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