Zeroooo solutions bc i did it earlier
<span>∠ABC ∠BAD is the correct answer
</span>
First list all the terms out.
e^ix = 1 + ix/1! + (ix)^2/2! + (ix)^3/3! ...
Then, we can expand them.
e^ix = 1 + ix/1! + i^2x^2/2! + i^3x^3/3!...
Then, we can use the rules of raising i to a power.
e^ix = 1 + ix - x^2/2! - ix^3/3!...
Then, we can sort all the real and imaginary terms.
e^ix = (1 - x^2/2!...) + i(x - x^3/3!...)
We can simplify this.
e^ix = cos x + i sin x
This is Euler's Formula.
What happens if we put in pi?
x = pi
e^i*pi = cos(pi) + i sin(pi)
cos(pi) = -1
i sin(pi) = 0
e^i*pi = -1 OR e^i*pi + 1 = 0
That is Euler's identity.
In the general case, it is (x, y+3), where y = f(x).
