Introduction:
In this question, we are given a complex number z and asked to find the angle between the vectors z and -iz. To solve this problem, we will first express the complex numbers in vector form and then use the dot product formula to find the angle between the vectors.

Expressing complex numbers in vector form:
A complex number z can be expressed in the form a + bi, where a and b are real numbers. We can also represent a complex number as a vector in the complex plane, with the real part representing the x-coordinate and the imaginary part representing the y-coordinate.

Let z = a + bi, then the vector representation of z is (a, b).

Similarly, the complex number -iz can be expressed as -ib + ia, which can be written as (-b, a) in vector form.

Finding the angle between vectors:
To find the angle between two vectors, we can use the dot product formula:

θ = arccos((u · v) / (|u| |v|))

where u and v are two vectors, · denotes the dot product, and |u| and |v| represent the magnitudes of the vectors.

In our case, the vectors u and v are z and -iz, respectively. Let's calculate the dot product and magnitudes:

u · v = (a, b) · (-b, a) = ab + ba = 2ab

|u| = √(a^2 + b^2)

|v| = √((-b)^2 + a^2) = √(b^2 + a^2)

Plugging these values into the formula, we get:

θ = arccos((2ab) / (√(a^2 + b^2) √(b^2 + a^2)))

Simplifying further:

θ = arccos(2ab / (a^2 + b^2))

Conclusion:
The angle between the vectors z and -iz is given by θ = arccos(2ab / (a^2 + b^2)). Therefore, the answer is d) none of these because the angle cannot be determined without knowing the values of a and b.

Pi