How exp(jwt)=cos(wt)+jsin(wt)?

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The discussion centers on understanding the equality exp(jwt) = cos(wt) + jsin(wt) and why the exponential function appears sinusoidal. Participants clarify that exp(ix) maps to the unit circle in the complex plane, which explains its periodic nature. They highlight the Taylor series expansion for both exponential and trigonometric functions, demonstrating how substituting ix into the series reveals the relationship between them. The confusion arises from the perception of exponential growth along the real axis versus its periodic behavior along the imaginary axis. Ultimately, the exponential function behaves differently in the complex domain, leading to sinusoidal outputs.
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Can anyone explain in simple terms on how they are equal? graphs are welcome.

The thing I don't understand is why exp(jwt) is sinusoidal. Wouldn't exp(jwt) go to infinity when t increases? I don't know how to understand the exp term as sinusoidal? in my mind they simply go to zero or infinity.
 
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Assuming j is the imaginary unit just compare the Taylor series for these two functions.
 
nomisme said:
Can anyone explain in simple terms on how they are equal? graphs are welcome.

The thing I don't understand is why exp(jwt) is sinusoidal. Wouldn't exp(jwt) go to infinity when t increases? I don't know how to understand the exp term as sinusoidal? in my mind they simply go to zero or infinity.

exp(x) behaves the way you expect. The image of exp(ix) is the unit circle on the complex plane.
 
notice they solve the same differential equation.
 
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I like Mathwonk's answer! A similar way to see that e^{ix}= cos(x)+ isin(x) is to use the Taylor's series at x= 0 (as jgens said):

e^{x}= 1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot
cos(x)= 1- x^2/2+ x^4/4!+ \cdot\cdot\cdot+ (-1)^n x^{2n}/(2n!)+ \cdot\cdot\cdot
sin(x)= x- x^3/3!+ x^5/5!+ \cdot\cdot\cdot+ (-1)^n x^{2n+1}/(2n+1)!+ \cdot\cdot\cdot

Each involves powers of x over the factorial of that power. Sine has only odd powers, cosine only even powers and sine and cosine have alternating sign.

Now see what happens when we replace "x" with "ix" in the series for e^x:
e^{ix}= 1+ (ix)+ (ix)^2/2+ (ix)^3/3!+ \cdot\cdot\cdot+ (ix)^n/n!+ \cdot\cdot\cdot
e^{ix}= 1+ ix+ i^2x^2/2+ i^3x^3/3!+ \cdot\cdot\cdot+ i^nx^n/n!+ \cdot\cdot\cdot
Of course, i^2= -1 so i^3= i^2(x)= -i, i^4= -i(i)= 1 and then it all repeats. That is, every odd power of ix is plus or minus ix while every even power is plus or minus 1. So
e^{ix}= 1+ ix- x^2/2- ix^3/3!+ \cdot\cdot\cdot+ (-1)^nx^{2n}/(2n)!+ (-1)^ni x^{2n+1}/(2n+1)!+ \cdot\cdot\cdot

and separating "real" and "imaginary" parts
e^{ix}= (1- x^2/2+ x^4/4!+ \cdot\cdot\cdot+ (-1)^n x^{2n}/(2n)!+ \cdot\cdot\cdot)+ i(x- x^3/3!+ \cdot\cdot\cdot+ (-1)^n x^{2n+1}/(2n+1)!)
e^{ix}= cos(x)+ isin(x)

I don't know how to understand the exp term as sinusoidal? in my mind they simply go to zero or infinity.
That's because you are used to dealing with real numbers. The exponential is increasing along along the real axis, periodic along the imaginary axis.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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