Question

If \(z = i\log(2 - \sqrt{3})\), then the value of \(\cos z\) will be:

• A. \(i\)
• B. \(2i\)
• C. 1
• D. 2

Hint:

\(\cos(ix) = \cosh x\), \(\sin(ix) = i\sinh x\).

Answer 1

The Correct Option is D

To solve the problem, we need to find the value of \(\cos z\) where \(z = i\log(2 - \sqrt{3})\).

Let's break this down step by step:

1. First, recognize that the logarithm of a number can be complex if the number is negative (or less than 1 in the case of a real base). The expression \(2 - \sqrt{3}\) is a positive real number, less than 1, which can potentially have a complex logarithm.
2. Since \(z = i\log(2 - \sqrt{3})\), start by letting \(w = \log(2 - \sqrt{3})\). Then, \(z = iw\).
3. To simplify this, let's first convert \(2 - \sqrt{3}\) to its polar form:
   • Let \(r = |2 - \sqrt{3}|\). Thus, \(r = \sqrt{(2)^2 + (-\sqrt{3})^2} = \sqrt{4 + 3} = \sqrt{7}\).
   • The angle, \(\theta\), for \(2 - \sqrt{3}\) is such that \(\cos\theta = \frac{2}{r}\) and \(\sin\theta = \frac{-\sqrt{3}}{r}\).
   • In this case, \(\cos\theta = \frac{2}{\sqrt{7}}\) and \(\sin\theta = \frac{-\sqrt{3}}{\sqrt{7}}\), driven from the Cartesian coordinates.
   • Upon finding \(\theta\), the logarithm in polar form is: \(\log(2 - \sqrt{3}) = \ln(\sqrt{7}) + i\theta\).
4. Substitute this into \(z = iw\):\[ z = i(\ln(\sqrt{7}) + i\theta) = i\ln(\sqrt{7}) - \theta. \]
5. Now, we must evaluate \(\cos z\), where \(z = i(w) = i\ln(\sqrt{7}) - \theta\).
6. The expression simplifies to \(\cos(i\ln(r)) = \cosh(\ln(\sqrt{7}))\).
7. By definition of \(\cosh\), this becomes:
   \[ \cosh(x) = \frac{e^x + e^{-x}}{2} \implies \cosh(\ln(\sqrt{7})) = \frac{\sqrt{7} + \frac{1}{\sqrt{7}}}{2} = \frac{7 + 1}{2 \cdot \sqrt{7}} = \frac{8}{2 \cdot \sqrt{7}} = 2. \]

Thus, the value of \(\cos z\) is \(\boxed{2}\).