Complex solution to cos x = 2?

[johann1301]

Are there any complex solutions to cos x = 2?

 

[Char. Limit]

Yes.

 

[Char. Limit]

Now for the next question. How does one find them, right? Well, you just need to take an alternate definition for the cosine function:

$$cos(x) = \frac{e^{i x} + e^{- i x}}{2}$$

Set the right side equal to 2, and solve for x. Hint, it's going to involve a quadratic equation in e^x.

 

[johann1301]

Thanks!

 

[CellCoree]

Yes, there are complex solutions to cos x = 2. In order to find these solutions, we can use the inverse cosine function (cos^-1) and the definition of complex numbers.

First, we can rewrite cos x = 2 as x = cos^-1(2). This means that x is the angle whose cosine is 2. However, in the real number system, the cosine function only takes values between -1 and 1. This means that there are no real solutions to cos x = 2.

To find the complex solutions, we can use the definition of complex numbers, where i is the imaginary unit (i^2 = -1). We can express 2 as 2 + 0i, and use the inverse cosine function to obtain x = cos^-1(2 + 0i).

Using the trigonometric identity cos(x + iy) = cos x cosh y - i sin x sinh y, we can expand cos x as cos x = cos x cosh 0 - i sin x sinh 0. Simplifying this, we get cos x = cos x - i sin x, which is the same as cos x = 2 + 0i.

Comparing the real and imaginary parts, we get cos x = 2 and -sin x = 0. This means that x must be an angle whose sine is 0, which can be any multiple of π. Therefore, the complex solutions to cos x = 2 are x = 2πn, where n is any integer.

In conclusion, there are infinitely many complex solutions to cos x = 2, given by x = 2πn, where n is any integer. These solutions can be expressed in the form of complex numbers as 2 + 0i.