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Question: Let \[m\] be a positive integer, then \[\log \left( { - mi} \right)\] A.\[\log m - i\pi \] B.\[\...

Let mm be a positive integer, then log⁡(−mi)\log \left( { - mi} \right)
A.log⁡m−iπ\log m - i\pi
B.log⁡m−iπ2\log m - i\dfrac{\pi }{2}
C.log⁡m+iπ\log m + i\pi
D.log⁡m+iπ2\log m + i\dfrac{\pi }{2}

Explanation

Solution

Here we will use the general form of the Euler form. Then we will apply the log to both sides of the Euler general form and solve it using the properties of the log function to get the value of the function log⁡(−mi)\log \left( { - mi} \right).

Complete step-by-step answer:
Given function is log⁡(−mi)\log \left( { - mi} \right).
Let the value inside the log function be zz i.e. z=−miz = - mi.
Therefore, the function becomes log⁡(z)\log \left( z \right).
Now we will use the basic of the Euler general form which is given by z=me−iπ/2z = m{e^{ - i\pi /2}}.
We will apply log to both of the Euler general form equations. Therefore we get
⇒log⁡(z)=log⁡(me−iπ/2)\Rightarrow \log \left( z \right) = \log \left( {m{e^{ - i\pi /2}}} \right)
Now we will use the basic property of the log function to solve the above equation.
Applying the logarithm property log⁡ab=log⁡a+log⁡b\log ab = \log a + \log b, we get
⇒log⁡(z)=log⁡(m)+log⁡(e−iπ/2)\Rightarrow \log \left( z \right) = \log \left( m \right) + \log \left( {{e^{ - i\pi /2}}} \right)
We know that when a log function is multiplied to the exponential function then it cancels out as they are the inverse function to each other. Therefore, we get
⇒log⁡(z)=log⁡(m)−iπ2\Rightarrow \log \left( z \right) = \log \left( m \right) - i\dfrac{\pi }{2}
Substituting the value of zz in the above equation, we get
⇒log⁡(−mi)=log⁡(m)−iπ2\Rightarrow \log \left( { - mi} \right) = \log \left( m \right) - i\dfrac{\pi }{2}
Hence, the value of log⁡(−mi)\log \left( { - mi} \right) is equal to log⁡(m)−iπ2\log \left( m \right) - i\dfrac{\pi }{2}.
So, option B is the correct option.

Note: We should know that the value inside the log function should never be zero or negative it should always be greater than zero. We should also remember that the value of the log⁡10\log 10 is equal to 1. The exponential function is the inverse function of the log function and vice versa which means that when a log function is multiplied to the exponential function, it cancels out and vice versa.