# 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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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, $$\displaystyle{\ln{{\left({i}\right)}}}={\ln{{\left({1}\right)}}}+{i}{\left({\frac{{\pi}}{{{2}}}}\right)}$$