The point doubling (Wiki link) requires Z^4 and when used with the double-and-add algorithm (needed for public point calculation, ECDH, and EC-based signatures) using $Z=1$ simplifies the calculation of Z^4 otherwise one may need 3 doublings. The double-and-add method where the res is not fixed to save time. Double-and-add Wikipedia version;
let bits = bit_representation(s) # the vector of bits (from MSB to LSB) representing s
let res = O # point at infinity
for bit in bits:
res = res + res # double
if bit == 1:
res = res + P # add
i = i - 1
return res
This is the beginning of the long story. The "m-fold double" (repeated doubling) calculates $[2^m]P$ and only calculates Z^4 once. When you need $[k]P$, you may need to represent $k$ in the binary form then use m-fold doubling when necessary. To have benefited from this, one has to calculate the cost before deciding to use m-fold double or not.
The answer is not easy and complete, since this one requires Z1=Z2 with 5M + 2S for addition Wiki version has 12M + 4S cost. The 5M + 2S has still Z1 multiplication and if Z1=1 that has zero cost.
In a short sentence, in general, Z1=1 simplifies the equations.
Since $(X_1:Y_1:Z_1)$ represents $(Z/Z^2,Y/Z^3)$ and $(X_1:Y_1:Z_1)$ is an equivalance relation that is $$(X_1:Y_1:Z_1) \sim (\lambda X_1:\lambda Y_1:\lambda Z_1)$$ one can simple convert the $Z_1 =1$ with $$(X_1/Z_1:Y_1/Z_1:1)$$
Keep in mind that 1/Z1 is not division, rather the inverse of Z1 on the defining field.
The Z1 on the other hand, is not staying there with Z1=1 under the operations. To benefit from this, one has to find the inverse and execute two multiplication. Finding inverse, on the other hand, is what we don't want since it is costly.
So, at least there is a benefit at the beginning of scalar multiplication.
