The faces of 0

Algebra Level 3

$\LARGE 0^{i^{2k}}$

For imaginary unit $i$ , the expression above is a number if:

Only if

k

is a non-negative even number Only if

k

is a positive even number Only if

k=0

k

is an even number

1 solution

Curtis Clement
Sep 23, 2015

Using $\ i^2 = -1$ $\ 0^{i^{2k}} = 0^{(-1)^k}$ This implies that k is even because 0 is a defined number whilst $\ 0^{(-1)} = \frac{1}{0}$ is not defined.

This implies that k should be non-negative i.e even numbers INCLUDING 0

Sourav Kumar Surya - 5 years, 8 months ago

Why must $k$ be non-negative? Negative even numbers also work. In fact, the solution set for $k$ is the set of all even integers (both non-negative and negative).

That is to say, the solution set for $k$ is $\{x:\frac x2\in\Bbb Z\}=\{\ldots,-4,-2,0,2,4,\ldots\}$ .

Prasun Biswas - 5 years, 8 months ago

why k should not equal to 0.

Sourav Kumar Surya - 5 years, 8 months ago

$0$ is also an even number. To be precise, the set of even numbers is described by $\{x:\frac x2\in\Bbb Z\}$ where $\Bbb Z$ is the set of all integers.

Note that $\frac{0}{2}=0\in\Bbb Z$ , hence $0$ is an even number by definition.

Prasun Biswas - 5 years, 8 months ago

The faces of 0

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