Question

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

Answer 1

Hint: 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 \[\log \left( { - mi} \right)\].

Complete step-by-step answer:
Given function is \[\log \left( { - mi} \right)\].
Let the value inside the log function be \[z\] i.e. \[z = - mi\].
Therefore, the function becomes \[\log \left( z \right)\].
Now we will use the basic of the Euler general form which is given by \[z = m{e^{ - i\pi /2}}\].
We will apply log to both of the Euler general form equations. Therefore we get
\[ \Rightarrow \log \left( z \right) = \log \left( {m{e^{ - i\pi /2}}} \right)\]
Now we will use the basic property of the log function to solve the above equation.
Applying the logarithm property \[\log ab = \log a + \log b\], we get
\[ \Rightarrow \log \left( z \right) = \log \left( m \right) + \log \left( {{e^{ - i\pi /2}}} \right)\]
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
\[ \Rightarrow \log \left( z \right) = \log \left( m \right) - i\dfrac{\pi }{2}\]
Substituting the value of \[z\] in the above equation, we get
\[ \Rightarrow \log \left( { - mi} \right) = \log \left( m \right) - i\dfrac{\pi }{2}\]
Hence, the value of \[\log \left( { - mi} \right)\] is equal to \[\log \left( m \right) - i\dfrac{\pi }{2}\].
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 \[\log 10\] 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.