Question

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

Answer 1

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.