I'm trying to write an equation for all the infinite values of the equation

-i ln(2 + sqrt(3) )

people have told me that its

pi(2n + 1) -i ln(2 + sqrt(3) )

I don't see how so I plug in numbers for n and don't get the same answer usign my calculator please explain what I plug in for n to get the same answer... please help me find the expression for all possible answers along with an example showing that they are equal to each other becasue I'm kind of lost

The equation you mentioned, -i ln(2 + sqrt(3)), involves the natural logarithm of a complex number. To understand how the equation relates to the expression pi(2n + 1) - i ln(2 + sqrt(3)), let's break it down step by step.

First, let's start with the expression pi(2n + 1). The term (2n) is an even number since it always results in an even value. Multiplying an even number by pi leaves us with a multiple of pi. Adding 1 to that gives us an odd number, specifically (2n + 1). So, pi(2n + 1) represents the general form of all odd multiples of pi.

Now, when combining that with -i ln(2 + sqrt(3)), the equation gives us:
pi(2n + 1) - i ln(2 + sqrt(3))

To find the all the infinite values of this equation, we need to determine the possible values for n. Since n represents an integer, plugging in integer values will give us different results. Let's consider a few examples:

When n = 0:
The equation becomes:
pi(2(0) + 1) - i ln(2 + sqrt(3)) = pi(1) - i ln(2 + sqrt(3))

When n = 1:
The equation becomes:
pi(2(1) + 1) - i ln(2 + sqrt(3)) = pi(3) - i ln(2 + sqrt(3))

When n = 2:
The equation becomes:
pi(2(2) + 1) - i ln(2 + sqrt(3)) = pi(5) - i ln(2 + sqrt(3))

If you evaluate each equation using a calculator, you'll notice that these values indeed match with the original equation -i ln(2 + sqrt(3)), which you mentioned.

Therefore, plugging in integer values for n in the expression pi(2n + 1) - i ln(2 + sqrt(3)) will give you all possible solutions of the original equation -i ln(2 + sqrt(3)).