ea + bi = ea(cos b + i sin b)
This formula gives rise to the equation eπi + 1 = 0. . Show how Euler’s formula can be used to derive this equation.
try getting the notation right:
e^(a+bi) = e^a (cosb + i sinb)
Now just plug in a=0, b=π
To derive Euler's formula, which states that e^(ix) = cos(x) + i sin(x), we can start by substituting π for x in the formula:
e^(iπ) = cos(π) + i sin(π)
Now, consider the values of cos(π) and sin(π):
cos(π) = -1
sin(π) = 0
Substituting these values into the equation, we have:
e^(iπ) = -1 + 0i
Since we know that e^(iπ) can be represented as -1 + 0i, we can rewrite the equation as:
e^(iπ) + 1 = 0
This is the derived equation from Euler's formula.