What is $ \ln ({i^2}) $ ?
Answer
279.3k+ views
Hint: The question is given as a logarithmic identity. The I here is the imaginary component which has the value $ \sqrt { - 1} $ .
To find out the solution, solve the bracket first and then come to the natural log.
Complete step-by-step answer:
As we know that,
Natural log: It is the logarithmic function which has the base equal to mathematical constant e.
i is the imaginary component of complex numbers.
I has the value = $ \sqrt { - 1} $
Given in the question,
= $ \ln ({i^2}) $
As
$
i = \sqrt { - 1} \\
{i^2} = ( - 1) \;
$
So the question becomes,
= $ \ln ( - 1) $
If we take complex number into considerations
$ \ln ({i^2}) = i\pi $
Otherwise undefined.
Note: Complex numbers play an important part in calculating ln(-1). If not for the consideration of complex number and rotation in complex plane the answer of the question is undefined. There is no real value exists
To find out the solution, solve the bracket first and then come to the natural log.
Complete step-by-step answer:
As we know that,
Natural log: It is the logarithmic function which has the base equal to mathematical constant e.
i is the imaginary component of complex numbers.
I has the value = $ \sqrt { - 1} $
Given in the question,
= $ \ln ({i^2}) $
As
$
i = \sqrt { - 1} \\
{i^2} = ( - 1) \;
$
So the question becomes,
= $ \ln ( - 1) $
If we take complex number into considerations
$ \ln ({i^2}) = i\pi $
Otherwise undefined.
Note: Complex numbers play an important part in calculating ln(-1). If not for the consideration of complex number and rotation in complex plane the answer of the question is undefined. There is no real value exists
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