Should not be forked from here as it is not about being smart (and smarter) with Python complex numbers.

Nice.

Complex part of the number.

What's this '1j'?

oh, when talking about byte-count, you where talking about the number of chars. 'Thought you were talking about the bytecode, actually. x)

Still longer :)

hypotenuse=lambda a,b:abs(a+b*1j)
from math import hypot as hypotenuse

Also this code contradicts the statements in the very first kumite about "let's go back to school, solve for hypotenuse", because you're simply importing a function which does everything for you (this also applies to @nithin5426 fork).

The whole point, of my fork was presenting a method which does not require any library imports.

Also this is the shortest byte-count you can possible achieve when searching for hypotenuse.

hypotenuse=lambda a,b:abs(a+b*1j)
    
import math;hypotenuse=lambda a,b:math.hypot(a,b)

This code is not smart, let alone smarter...