# What is ln(i^2) ?

##### 2 Answers
Nov 24, 2017

$\ln \left({i}^{2}\right) = \pi i$

#### Explanation:

$\ln \left(i\right) = \ln \left(\sqrt{- 1}\right) = \ln \left({\left(- 1\right)}^{\frac{1}{2}}\right) = \frac{1}{2} \ln \left(- 1\right) = \frac{1}{2} \cdot \pi i$

so,
$\ln \left({i}^{2}\right) = 2 \cdot \ln \left(i\right) = 2 \cdot \frac{1}{2} \cdot \pi i = \pi i$

Nov 24, 2017

$\ln \left({i}^{2}\right) = \ln \left(- 1\right) = \pi i$

#### Explanation:

The function ${e}^{x}$ considered as a real valued function of real numbers is one to one from $\left(- \infty , \infty\right)$ onto $\left(0 , \infty\right)$

Therefore the real valued logarithm is a well defined function from $\left(0 , \infty\right)$ onto $\left(- \infty , \infty\right)$

The function ${e}^{z}$ considered as a complex valued function of complex numbers is many to one from $\mathbb{C}$ onto $\mathbb{C} \text{\} \left\{0\right\}$.

In particular, note that ${e}^{2 \pi i} = 1$. So if ${e}^{z} = c$ then ${e}^{z + 2 n \pi i} = c$ for any integer $n$.

Therefore the complex logarithm has multiple branches, with a principal branch from $\mathbb{C} \text{\} \left\{0\right\}$ onto $\left\{x + y i \in \mathbb{C} : y \in \left(- \pi , \pi\right]\right\}$ (using the normal range of $A r g \left(z\right) \in \left(- \pi , \pi\right]$)

The expression $\ln \left(z\right)$ denotes this principal value.

So whereas $z = 7 i \pi$ is a root of ${e}^{z} = - 1$, it is not the principal value of $\ln \left({i}^{2}\right) = \ln \left(- 1\right)$.

The principal value is $\ln \left(- 1\right) = \pi i$

In general, we can write a formula for the principal value of the logarithm of a complex number $z$ as:

$\ln z = \ln \left\mid z \right\mid + A r g \left(z\right) i$