GCD of two whole complex numbers GCD(5-3i,7+i) in Z[i]

Use factorizations $5-3i=(1-i)(4+i)$ and $7+i = (1-i)(3+4i)$. Note that $(4+i)(4-i)=17$ and $(3+4i)(3-4i)=25$ are coprime over $\mathbb Z$, hence $\mathbb Z[i]$ as well. Therefore their gcd is $(1-i)$ up to a unit.

If you want a more simpler solution, check the solution by @Just a user.

$f(5 - 3i) = 5^2 + 3^2 = 25 + 9 = 34$, and $f(7 + i) = 7^2 + 1^2 = 50$
Note that $GCD(34, 50) = 2$, thus any common divisor of the aforementioned Gaussian integers must also divide 2.

Since 2 can be factorized into $(1 + i)(1 - i)$, and $(1 + i) = i(1 - i),$ thus $GCD(5 - 3i, 7 + i) = (1 - i) (or \:\:1 + i)$.

This is clearly true, since: $$(5 - 3i) = (4 - i^2) +(i-4i) = (1 - i)(4 + i),$$ and $$(7 + i) = (3 - 4i^2)+(4i-3i)=(1 - i)(3 + 4i)$$