Question

Find the general and principal value of \[\log \left( { - 1 + i} \right) - \log \left( { - 1 - i} \right)\] ?

Answer 1

Hint: In the above given question, we are given an expression written as \[\log \left( { - 1 + i} \right) - \log \left( { - 1 - i} \right)\] . This expression consists of the logarithmic function and the complex number unit \[i\] , called iota defined as \[i = \sqrt { - 1} \] . We have to find the general and principal value of the given expression.

Complete answer:
Given expression is,
\[ \Rightarrow \log \left( { - 1 + i} \right) - \log \left( { - 1 - i} \right)\]
We have to find the general and principal value of the given expression.
First let us rewrite the given expression in a smaller form.
Using the logarithmic formula of division that is given by \[\log A - \log B = \log \dfrac{A}{B}\] , we can write
\[ \Rightarrow \log \left( { - 1 + i} \right) - \log \left( { - 1 - i} \right) = \log \dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 - i} \right)}}\]
Now consider \[\dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 - i} \right)}}\] .
Hence, multiplying and dividing this expression with the conjugate of the denominator, that is \[ - 1 + i\] , we get
\[ \Rightarrow \dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 - i} \right)}} = \dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 - i} \right)}} \cdot \dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 + i} \right)}}\]
That gives us,
\[ \Rightarrow \dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 - i} \right)}} = \dfrac{{{{\left( { - 1 + i} \right)}^2}}}{{1 - {i^2}}}\]
Or,
\[ \Rightarrow \dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 - i} \right)}} = \dfrac{{1 + {i^2} - 2i}}{{1 - {i^2}}}\]
We can also write is as,
\[ \Rightarrow \dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 - i} \right)}} = \dfrac{{1 - 1 - 2i}}{{1 - \left( { - 1} \right)}}\]
That gives us,
\[ \Rightarrow \dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 - i} \right)}} = \dfrac{{ - 2i}}{2}\]
Hence,
\[ \Rightarrow \dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 - i} \right)}} = - i\]
Therefore we have the original expression as,
\[ \Rightarrow \log \left( { - 1 + i} \right) - \log \left( { - 1 - i} \right) = \log \dfrac{{\left( { - 1 + i} \right)}}{{\left( { - 1 - i} \right)}} = \log \left( { - i} \right)\]
Now we have to find the value of \[\log \left( { - i} \right)\] .
Since we know that \[{e^{i\theta }} = \cos \theta + i\sin \theta \] ,
Therefore, taking \[\theta = \dfrac{{3\pi }}{2}\] we can write
\[ \Rightarrow {e^{i\dfrac{{3\pi }}{2}}} = \cos \dfrac{{3\pi }}{2} + i\sin \dfrac{{3\pi }}{2}\]
That gives us,
\[ \Rightarrow {e^{i\dfrac{{3\pi }}{2}}} = \cos \left( {\pi + \dfrac{\pi }{2}} \right) + i\sin \left( {\pi + \dfrac{\pi }{2}} \right)\]
Or,
\[ \Rightarrow {e^{i\dfrac{{3\pi }}{2}}} = - \cos \dfrac{\pi }{2} - i\sin \dfrac{\pi }{2}\]
Now since, \[\cos \dfrac{\pi }{2} = 0\] and \[\sin \dfrac{\pi }{2} = 1\]
Hence, we have
\[ \Rightarrow {e^{i\dfrac{{3\pi }}{2}}} = - i\]
Therefore, now we have the original expression as,
\[ \Rightarrow \log \left( { - 1 + i} \right) - \log \left( { - 1 - i} \right) = \log \left( { - i} \right) = \log \left( {{e^{i\dfrac{{3\pi }}{2}}}} \right)\]
Now, we can write
\[ \Rightarrow \log \left( {{e^{i\dfrac{{3\pi }}{2}}}} \right) = i\dfrac{{3\pi }}{2}\log e\]
For a natural logarithmic function, we have \[{\log _e}e = 1\] .
Hence, that gives us
\[ \Rightarrow \log \left( {{e^{i\dfrac{{3\pi }}{2}}}} \right) = i\dfrac{{3\pi }}{2}\]
Therefore, the principal value of \[\log \left( { - 1 + i} \right) - \log \left( { - 1 - i} \right)\] is \[i\dfrac{{3\pi }}{2}\] .
Now, for the general value of this expression we can write \[\theta \] as,
\[ \Rightarrow \theta = \dfrac{{3\pi }}{2} + 2k\pi \]
Where \[k \in \mathbb{Z}\] i.e. \[k\] is any integer.
Therefore, the general value of \[\log \left( { - 1 + i} \right) - \log \left( { - 1 - i} \right)\] is \[i\left( {\dfrac{{3\pi }}{2} + 2k\pi } \right)\] .

Note:
If the trigonometric or complex equation involves the angle \[\theta \] such that \[0 \leqslant \theta \leqslant 2\pi \] , then the solutions are called principal solutions.
Whereas a general solution is the one which involves the integer \[k\] and gives all the solutions of that trigonometric equation. Here, the symbol \[\mathbb{Z}\] is used to denote the set of integers.