How do I express the math equation i + e^(i * pi) in Python?

You can express the math equation i + e^(i * pi) in Python using the cmath module, which provides support for complex numbers.

import cmath

# Define i as the imaginary unit
i = 1j

# Calculate the expression i + e^(i * pi)
result = i + cmath.exp(i * cmath.pi)

print(result)

• 1j represents the imaginary unit 𝑖
• cmath.exp() calculates the exponential function for complex numbers
• cmath.pi is the constant π

Yes, your use of j is incorrect. Use complex() to represent complex numbers rather than trying to calculate them with sqrt().

import cmath
complex(0, 1) + cmath.exp(complex(0, cmath.pi))

(-1+1.0000000000000002j)

From the help() output of complex()

class complex(object)
 |  complex(real=0, imag=0)
 |  
 |  Create a complex number from a real part and an optional imaginary part.
 |  
 |  This is equivalent to (real + imag*1j) where imag defaults to 0.

As @mkrieger1 points out in the comments, a simpler solution is

1j + cmath.exp(1j * cmath.pi)

Since, as shown in these answers, cmath.exp(1j*w) is broken, at least for w = cmath.pi, the math fact that e^(jw) = cos(w) +jsin(w) can be used instead. Testing has shown that cmath.sin(w) is correct for w in the range of -cmath.pi/2 through cmath.pi/2. Since testing has shown that cmath.cos(w) is not correct for any range of w of size pi/2, the math fact that cos w == sin(w+pi/2) should be used. The math fact that sin(w - pi) = -sin(w) is also needed. The above, altogether:

import cmath

print("i + e^(i * pi) = ", end = '')

# 1j + cmath.exp(1j * cmath.pi) ==
# 1j + cmath.cos(cmath.pi) + 1j * cmath.sin(cmath.pi) == 
# 1j + cmath.sin(cmath.pi + cmath.pi/2) + 1j * cmath.sin(cmath.pi) == 
# 1j + cmath.sin(1.5 * cmath.pi) + 1j * cmath.sin(cmath.pi) ==
# 1j + cmath.sin(1.5 * cmath.pi) - 1j * cmath.sin(cmath.pi - cmath.pi) ==
# 1j + cmath.sin(1.5 * cmath.pi) - 1j * cmath.sin(0) ==
# 1j - cmath.sin(1.5 * cmath.pi - cmath.pi) - 1j * cmath.sin(0) ==
# 1j - cmath.sin(cmath.pi/2) - 1j * cmath.sin(0) ==

print("i + e^(i * pi) = ", end = '')
print(1j - cmath.sin(cmath.pi/2) - 1j * cmath.sin(0))