Microsoft Q# : problems with rotate function R() - q#

Question

Recently I've been working on algorithms of quantum computing where I faced some problems.
As there are probabilities for qubits to be observed as |1> or |0>, the rotation function is supposed to change the status of each qubit so that there can be more or fewer chances for them to be measured in a specific status.
In my program, I applied Hadamard transformation to a qubit and then rotated it with an angle θ(90°) in the z-axis, which was to make an equal probability of |0> and |1> measurement and then the rotation made the probs of status |1> 100%, but it didn't work anyway. The probs for |1> will always be 50% which is not supposed to be.
Here is my code:
operation Testing () : (Int)
{
body
{
mutable result = -1;
using (qubit = Qubit[1])
{
H(qubit[0]);
R(PauliZ, PI() / -2.0, qubit[0]);
//assertprob is a function to help test the probabilities of qubits in specific status
AssertProb([PauliZ], qubit, One, 0.5,"Measuring in conjugate basis did not give 50/50 results.", 1e-5);
//AssertProb([PauliZ], qubit, Zero, 0.5,"Measuring in conjugate basis did not give 50/50 results.", 1e-5);
let value = M(qubit[0]);
if(value == One)
{
set result = 1;
} else {
set result = 0;
}
ResetAll(qubit);
}
return result;
}
}
A photo from Microsoft Q# documentation may help explain the sphere structure. After the Hadamard transformation, the qubit is in status [1/√2,1/√2] and then the rotation of -π/2 makes it to status [0,1].
(source: microsoft.com)

Answer 1

Applying a rotation around the Pauli Z axis doesn't change the measurement probabilities of Z-axis measurements. You can visualize it this way:
Your qubit's initial state is at +1 on the Z axis.
The H operation moves it onto +1 on the X axis.
Any rotation around the Z axis keeps the qubit's state in the X-Y plane, which is equally likely to measure as 0 or as 1 in the Z axis.
A pi/2 or 3pi/2 rotation around the Y axis will get the state back to the Z axis.