If 1 + i is a solution of z^44 + 4 = 0, if we substitute z with 1 + i, we should get 0.

Doing the same

(1 + i)^4 + 4

=> [(1 + i)^2]^2 + 4

=> [1 + i^2 + 2i]^2 +4

=> [1 - 1 + 2i]^2 + 4

=> (2*i)^2 + 4

=> 4* i^2 + 4

=> -4 + 4

=> 0

So we arrive at 0.

**Therefore we can say that 1 + i is a solution of z^4 + 4 = 0.**