Question

If $z=i\left(\log \left(2-\sqrt{3}\right)\right)$, then $\cos z$ is equal to

• A. $i$
• B. $2i$
• C. $1$
• D. $2$

Answer 1

The correct option is D

$2$

Explanation for the correct option.

Find the value of $\cos z$.

It is known that: $\cos z=\frac{e^{iz}+e^{-iz}}{2}$.

Now it is given that $z=i\log \left(2-\sqrt{3}\right)$, so substitute $i\log \left(2-\sqrt{3}\right)$ for $z$ in the identity.

$\begin{array}{rcl}\cos \left(i\log \left(2-\sqrt{3}\right)\right)& =& \frac{e^{i·i\log \left(2-\sqrt{3}\right)}+e^{-i·i\log \left(2-\sqrt{3}\right)}}{2}\\ & =& \frac{e^{i^2\log \left(2-\sqrt{3}\right)}+e^{-i^2\log \left(2-\sqrt{3}\right)}}{2}\\ & =& \frac{e^{-\log \left(2-\sqrt{3}\right)}+e^{-(-1)\log \left(2-\sqrt{3}\right)}}{2}\left[i=-1\right]\\ & =& \frac{e^{\log \left({\displaystyle \frac{1}{2-\sqrt{3}}}\right)}+e^{\log \left(2-\sqrt{3}\right)}}{2}\left[-\log \left(m\right)=\log \left(\frac{1}{m}\right)\right]\\ & =& \frac{{\displaystyle \frac{1}{2-\sqrt{3}}}+2-\sqrt{3}}{2}\left[a^{{\log }_a\left(b\right)}=b\right]\end{array}$

On further simplification,

$\begin{array}{rcl}\frac{{\displaystyle \frac{1}{2-\sqrt{3}}+2-\sqrt{3}}}{2}& =& \frac{{\displaystyle \frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2-\sqrt{3}}}{2}\\ & =& \frac{{\displaystyle \frac{2+\sqrt{3}}{4-3}+2-\sqrt{3}}}{2}\\ & =& \frac{2+\sqrt{3}+2-\sqrt{3}}{2}\\ & =& \frac{4}{2}\\ & =& 2\end{array}$

So, the value of $\cos z$ is $2$.

Hence, the correct option is D.