Understanding Euler's Formula for Complex Numbers - Tips for Evaluating e^ix

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Euler's formula, e^ix = cosx + isinx, connects complex numbers in polar form to their rectangular representation. To evaluate expressions like e^ix, one can utilize Taylor series for exponential, sine, and cosine functions, demonstrating the relationship between these functions. The general form re^{iθ} represents complex numbers, where r is the radius and θ is the angle, facilitating conversions between polar and rectangular coordinates. Evaluating expressions with imaginary exponents, such as 3^{2i}, can also be accomplished using Euler's formula. Understanding these concepts clarifies the evaluation of complex numbers and their properties.
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I just started studying complex numbers. It says complex numbers in polar form can be expressed as a power of e:

e^ix = cosx + isinx

I don't quite understand how this equation works.
How do i evaluate e^ix? And how does taking e to the power of ix get me a complex number a + bi or even in polar form?
 
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To get all complex numbers, you need the somewhat more general form:
re^{i\theta}=r\sin{\theta}+ir\cos{\theta}

Now, the other direction is also not bad
z=x+iy=\sqrt{x^2+y^2} e^{i\arctan{\frac{x}{y}}}

If you think of r as the radius, and \theta as the angle anti-clockwise from the postive axis, you'll find that this corresponds to switching between polar and rectangular coordinates.
 
So re^{i\theta} is merely a convenience in writing?
 
It is more than a convenience. One way to look at it is by power series of eix and compare to power series for sinx and cos x. You will see that eix=cosx+isinx.
 
Ok, yes i think I'm beginning to understand. How do one evaluate xi? For example, what's 32i?
 
If you are thinking of x^n being the product of n x's, and want to apply that here, you are out of luck. When the exponent is imaginary, it doesn't really carry over. Think of it as a generalization.

Are you familiar with Taylor series?

e^x=1+x+{x^2\over 2}+{x^3\over 3!}+{x^4\over 4!}+...

\sin x=x-{x^3\over 3!}+{x^5\over 5!}-...

\cos x=1-{x^2\over 2}+{x^4\over 4!}-...

The result follows from these.
 
Well, you can do the following:

3^{2i}=e^{\ln{3} *2i}=\cos{2\ln{3}}+i\sin{2\ln{3}}

If you're feeling a bit more adventerous, there is a power series for 3^x which is going to give you the same result.
 
Ok, thanks! So the only way of evaluating xiy is by using Euler's formula, which we know to be true? I can see the use of the formula more clearly now.
 
For the record, x^y is usually a multivalued function.
 
  • #10
Originally posted by AndersHermansson
Ok, thanks! So the only way of evaluating xiy is by using Euler's formula, which we know to be true? I can see the use of the formula more clearly now.
Think of it as a way to calculate x^{iy}.
 
  • #11
Originally posted by NateTG
To get all complex numbers, you need the somewhat more general form:
re^{i\theta}=r\sin{\theta}+ir\cos{\theta}

Now, the other direction is also not bad
z=x+iy=\sqrt{x^2+y^2} e^{i\arctan{\frac{x}{y}}}

If you think of r as the radius, and \theta as the angle anti-clockwise from the postive axis, you'll find that this corresponds to switching between polar and rectangular coordinates.

should that be arctan(y/x)?
 
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