Let z be any complex number in polar form with 0 <= theta<2 pi. Use z=re^{i theta} to define the logarithm of a complex number. (a)Find ln(-1)

Daniaal Sanchez 2021-09-09 Answered

Let z be any complex number in polar form with \(\displaystyle{0}\leq\theta{<}{2}\pi\). Use \(\displaystyle{z}={r}{e}^{{{i}\theta}}\) to define the logarithm of a complex number.
(a)Find \(\displaystyle{\ln{{\left(-{1}\right)}}}\)
(b)Find \(\displaystyle{\ln{{\left({i}\right)}}}\)

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Expert Answer

Cristiano Sears
Answered 2021-09-10 Author has 28669 answers

Step 1
When \(\displaystyle{z}={r}{e}^{{{i}\theta}}\) then
\(\displaystyle{\ln{{\left({z}\right)}}}={\ln{{\left({r}\right)}}}+{i}{\left(\theta+{2}\pi{k}\right)}\)
where k is integer
(a) \(\displaystyle{\ln{{\left(-{1}\right)}}}\)
\(\displaystyle{z}=-{1}\)
\(\displaystyle\Rightarrow{z}=-{1}+{0}{i}\)
where it is of the form z=x+iy
then \(\displaystyle{r}=\sqrt{{{x}^{{2}}+{y}^{{2}}}}=\sqrt{{{\left({1}\right)}^{{2}}+{0}^{{2}}}}={1}\)
\(\displaystyle\theta={{\tan}^{{-{1}}}{\left({\frac{{{y}}}{{{x}}}}\right)}}={{\tan}^{{-{1}}}{\left({\frac{{{0}}}{{-{1}}}}\right)}}={{\tan}^{{-{1}}}{\left({0}\right)}}={0}\)
In polar form z can be written as
\(\displaystyle{z}={r}{e}^{{{i}\theta}}{\left(\because{0}\leq\theta{<}{2}\pi\right)}\)
\(\displaystyle={i}{e}^{{{i}\theta}}\)
then \(\displaystyle{\ln{{\left(-{1}\right)}}}={\ln{{\left({1}\right)}}}+{i}{\left({0}\right)}\)
\(\displaystyle{\ln{{\left(-{1}\right)}}}={\ln{{\left({1}\right)}}}\)
(b) \(\displaystyle{\ln{{\left({i}\right)}}}\)
\(\displaystyle{z}={i}\)
\(\displaystyle\Rightarrow{z}={0}+{1}{i}\ \text{ where }\ {x}={0},{y}={1}\)
then \(\displaystyle{r}=\sqrt{{{\left({0}\right)}^{{2}}+{\left({1}\right)}^{{2}}}}={1}\)
\(\displaystyle\theta={{\tan}^{{-{1}}}{\left({\frac{{{y}}}{{{x}}}}\right)}}={{\tan}^{{-{1}}}{\left({\frac{{{1}}}{{{0}}}}\right)}}={\frac{{\pi}}{{{2}}}}\)
In polar form
\(\displaystyle{z}={r}{e}^{{{i}\theta}}\)
substitute r=1 , \(\displaystyle\theta={\frac{{\pi}}{{{2}}}}\)
\(\displaystyle{z}={e}^{{{i}{\frac{{\pi}}{{{2}}}}}}\)
then logarithm of complex number
\(\displaystyle{z}={e}^{{{i}{\frac{{\pi}}{{{2}}}}}}\) is
\(\displaystyle{\left(\because\leq\theta{<}{2}\pi\right)}\)
\(\displaystyle{\ln{{\left({1}\right)}}}+{i}{\left({\frac{{\pi}}{{{2}}}}\right)}\)
Thus, \(\displaystyle{\ln{{\left({i}\right)}}}={\ln{{\left({1}\right)}}}+{i}{\left({\frac{{\pi}}{{{2}}}}\right)}\)

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