Question

If $z=i\log \left( 2-\sqrt{3} \right)$, then $\cos z=$
(a) i
(b) 2i
(c) 1
(d) 2

Answer 1

At first consider the formula $\cos z=\dfrac{{{e}^{iz}}+{{e}^{-iz}}}{2}$, where z is equal to $i\log \left( 2-\sqrt{3} \right)$ and then use the fact that value of ${{i}^{2}}=-1$ after that put the identity that ${{e}^{\log a}}$ is a and get what is asked.

Complete step-by-step answer:
In the question we are given a complex number z whose value is $z=i\log \left( 2-\sqrt{3} \right)$ and we have to find the value of $\cos z$.
Before doing so, we will learn what complex numbers are.
A complex number is a number that can be written in the form of a + bi, where a, b are real numbers and i is a solution of the equation ${{x}^{2}}=-1$. This is because no real value satisfies for equation ${{x}^{2}}+1=0$ or ${{x}^{2}}=-1$, hence i is called an imaginary number. For the complex number a + ib, a is considered as real part and b as imaginary part.
Despite the historical nomenclature “imaginary”, complex numbers are regarded in the mathematical sciences as just as “real” as real numbers and are fundamental in any aspect of scientific description of the natural world.
Now as we know that value of $\cos z$ can be represented as,
$\cos z=\dfrac{{{e}^{iz}}+{{e}^{-iz}}}{2}$
So now instead of z, let’s write $i\log \left( 2-\sqrt{3} \right)$ so we get,
$\cos \left( i\log \left( 2-\sqrt{3} \right) \right)=\dfrac{{{e}^{i\left( i\log \left( 2-\sqrt{3} \right) \right)}}+{{e}^{-i\left( i\log \left( 2-\sqrt{3} \right) \right)}}}{2}$
So we can write it as,
$\cos \left( i\log \left( 2-\sqrt{3} \right) \right)=\dfrac{{{e}^{{{i}^{2}}\log \left( 2-\sqrt{3} \right)}}+{{e}^{-{{i}^{2}}\log \left( 2-\sqrt{3} \right)}}}{2}$
Now we will use the fact that value of ${{i}^{2}}=-1$ so we get,
$\cos \left( i\log \left( 2-\sqrt{3} \right) \right)=\dfrac{{{e}^{-\log \left( 2-\sqrt{3} \right)}}+{{e}^{\log \left( 2-\sqrt{3} \right)}}}{2}$
So, we have $\dfrac{{{e}^{\log \dfrac{1}{2-\sqrt{3}}}}+{{e}^{\log \left( 2-\sqrt{3} \right)}}}{2}$
Now we will apply the identity that ${{e}^{\log a}}$ is equal to a.
Applying this, we get ${{e}^{\log \dfrac{1}{2-\sqrt{3}}}}=\dfrac{1}{2-\sqrt{3}}$ and ${{e}^{\log 2-\sqrt{3}}}=2-\sqrt{3}$.

So, we can write it as $\cos \left( i\log \left( 2-\sqrt{3} \right) \right)=\dfrac{\dfrac{1}{2-\sqrt{3}}+2-\sqrt{3}}{2}$
Now at first rationalize $\dfrac{1}{2-\sqrt{3}}$by multiplying it with $2+\sqrt{3}$ we get,
$\dfrac{1}{2-\sqrt{3}}\times \dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$
So we get,
$\cos \left( i\log \left( 2-\sqrt{3} \right) \right)=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{2}$
Simplifying it further, we will get $\cos \left( i\log \left( 2-\sqrt{3} \right) \right)=2$

So, the correct answer is “Option d”.

Note: Generally while solving students can miss out certain identities such as ${{e}^{\log a}}$ equal to a or fact that value of ${{i}^{2}}$ is equal to -1. Also they can make mistakes while rationalizing which can make their answers wrong so they should be careful about that.