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Question: If \(z = i\log (2 - \sqrt 3 )\) , then \(\cos z\) is equal to \(1)i\) \(2)2i\) \(3)1\) \(4)2...

If z=ilog⁡(2−3)z = i\log (2 - \sqrt 3 ) , then cos⁡z\cos z is equal to
1)i1)i
2)2i2)2i
3)13)1
4)24)2

Explanation

Solution

First, complex numbers are the real and imaginary combined numbers as in the form of z=x+iyz = x + iy, where x and y are the real numbers and ii is the imaginary.
Imaginary ii can be also represented into the real values only if, i2=−1{i^2} = - 1
Since we have given the value of the z, then we need to find the value of cos⁡z\cos z using the given information.

Formula used: cos⁡z=eiz+e−iz2\cos z = \dfrac{{{e^{iz}} + {e^{ - iz}}}}{2}

Complete step-by-step solution:
Since from the given that we have z=ilog⁡(2−3)z = i\log (2 - \sqrt 3 ) also the trigonometry cosine can be expressed as the exponent of cos⁡z=eiz+e−iz2\cos z = \dfrac{{{e^{iz}} + {e^{ - iz}}}}{2}
Now just substitute the value of the z, in the cosine then we have cos⁡z=eiz+e−iz2⇒cos⁡z=ei(ilog⁡(2−3))+e−i(ilog⁡(2−3))2\cos z = \dfrac{{{e^{iz}} + {e^{ - iz}}}}{2} \Rightarrow \cos z = \dfrac{{{e^{i(i\log (2 - \sqrt 3 ))}} + {e^{ - i(i\log (2 - \sqrt 3 ))}}}}{2}
Since we know that i2=−1{i^2} = - 1 then we get cos⁡z=ei(ilog⁡(2−3))+e−i(ilog⁡(2−3))2⇒cos⁡z=e−log⁡(2−3)+elog⁡(2−3)2\cos z = \dfrac{{{e^{i(i\log (2 - \sqrt 3 ))}} + {e^{ - i(i\log (2 - \sqrt 3 ))}}}}{2} \Rightarrow \cos z = \dfrac{{{e^{ - \log (2 - \sqrt 3 )}} + {e^{\log (2 - \sqrt 3 )}}}}{2}
Since exponent and the logarithm is the inverse process and they can be represented as elog⁡(a)=a,e−log⁡(a)=1a{e^{\log (a)}} = a,{e^{ - \log (a)}} = \dfrac{1}{a}
Thus, by this concept, we have the above equation as cos⁡z=e−log⁡(2−3)+elog⁡(2−3)2⇒cos⁡z=1(2−3)+(2−3)2\cos z = \dfrac{{{e^{ - \log (2 - \sqrt 3 )}} + {e^{\log (2 - \sqrt 3 )}}}}{2} \Rightarrow \cos z = \dfrac{{\dfrac{1}{{(2 - \sqrt 3 )}} + (2 - \sqrt 3 )}}{2}
Now cross multiplying we get cos⁡z=1+(2−3)2(2−3)2\cos z = \dfrac{{\dfrac{{1 + {{(2 - \sqrt 3 )}^2}}}{{(2 - \sqrt 3 )}}}}{2} and further soling we have cos⁡z=1+(2−3)2(2−3)2⇒cos⁡z=4+4−43(2−3)2⇒4((2−3))(2−3)×12\cos z = \dfrac{{\dfrac{{1 + {{(2 - \sqrt 3 )}^2}}}{{(2 - \sqrt 3 )}}}}{2} \Rightarrow \cos z = \dfrac{{\dfrac{{4 + 4 - 4\sqrt 3 }}{{(2 - \sqrt 3 )}}}}{2} \Rightarrow \dfrac{{4((2 - \sqrt 3 ))}}{{(2 - \sqrt 3 )}} \times \dfrac{1}{2}
Hence canceling the common terms, we have cos⁡z=4((2−3))(2−3)×12⇒cos⁡z=4×12⇒cos⁡z=2\cos z = \dfrac{{4((2 - \sqrt 3 ))}}{{(2 - \sqrt 3 )}} \times \dfrac{1}{2} \Rightarrow \cos z = 4 \times \dfrac{1}{2} \Rightarrow \cos z = 2
Therefore, the option 4)24)2 is correct.

Note: We will first understand what the logarithmic operator represents in mathematics. A logarithm function or log operator is used when we have to deal with the powers of a number, to understand it better which is log⁡xm=mlog⁡x\log {x^m} = m\log x
The conjugate of a complex number represents the reflection of that complex number about the real axis on the argand plane.
When the imaginary ii of the complex number is replaced with −i - i , we get the conjugate of that complex number that shows the image of the particular complex number about the plane.
The logarithm function we used log⁡xm=mlog⁡x\log {x^m} = m\log x and logarithm derivative function can be represented as ddxlog⁡x=1x\dfrac{d}{{dx}}\log x = \dfrac{1}{x}
And the exponent of the logarithm is elog⁡(a)=a,e−log⁡(a)=1a{e^{\log (a)}} = a,{e^{ - \log (a)}} = \dfrac{1}{a}