Given that an equation of the form f(x) = x^n has n solutions, and that the solutions are equally spaced at a given radius in the complex plane, discuss why complex solutions must occur in conjugate pairs when .

because one solution will lie on the positive real axis. The rest are evenly spaced above and below that axis, giving conjugate pairs.

To discuss why complex solutions must occur in conjugate pairs when n is odd, we first need to understand the nature of complex numbers and their representation in the complex plane.

Complex numbers are numbers of the form a + bi, where a and b are real numbers and i is the imaginary unit (i.e., √(-1)). When we plot complex numbers on a two-dimensional plane, we refer to it as the complex plane. The real part is represented on the x-axis, and the imaginary part on the y-axis.

Now let's consider the equation f(x) = x^n. For simplicity, let's assume n is an odd positive integer, such as 3 or 5.

When we solve this equation, we are looking for values of x that make f(x) equal to zero. In other words, we want to find the values of x that satisfy x^n = 0.

For any odd value of n, there will always be at least one real solution (x = 0) because any real number raised to an odd power will result in zero if the number itself is zero. This real solution will be present regardless of whether n is odd or even.

However, when n is odd, there will be additional complex solutions that occur in conjugate pairs. This is because the complex solutions arise from the imaginary roots of the equation.

To understand why complex solutions occur in conjugate pairs, consider the equation x^n = 0. Suppose z is a complex solution of the equation, where z = a + bi, with a and b being real numbers.

If we substitute z into the equation, we get (a + bi)^n = 0. Expanding this using the binomial theorem, we obtain a sum of terms, each of which contains a power of a and a power of b.

Since n is odd, there will be at least one term in the expansion that contains an odd power of i. This is because raising i to any odd power yields either i, -i, or -1.

Now, if z = a + bi is a solution, then its conjugate z* = a - bi will also be a solution. This is due to the property that when we take the complex conjugate of z^n, we get the same result as taking the complex conjugate of z, raised to the power n.

Therefore, if z is a complex solution, then z* must also be a solution, and vice versa. These solutions will be reflections of each other across the real axis in the complex plane.

Consequently, when n is odd, the equation f(x) = x^n will have one real solution and an equal number of complex solutions occurring in conjugate pairs in the complex plane.

In summary, when n is an odd positive integer, the complex solutions of the equation f(x) = x^n occur in conjugate pairs in the complex plane due to the nature of complex numbers and the odd powers of i.