How is $i^3$ = $\sqrt{-i}$.

Question

I new to complex numbers.So please don’t mind.

We know $i^2$ = -1.

Then $i^3$ is written as $= -i$

Then does it’s steps of writing follow this:

$-1^2$= $i^3$= 1

So in terms i = It says in my book

= - $\sqrt{-i}$.

How did we get this ?

I have done calculation mistake while writing this question.I have now answered this question.

Certainly you must be mistaking $\sqrt{-1}$ for $\sqrt{-i}$. — Andrew Chin, Dec 19 '20 at 09:10

score 0 · Accepted Answer · answered Dec 19 '20 at 10:00

$i^3$ = $\sqrt{-1}$. * $\sqrt{-1}$.* $\sqrt{-1}$.

Therefore it equals = -1 * $\sqrt{-1}$.

Or we can write it also using the property he listed in his question.

$i^2$ * i = -i.

Remember:When a and b both are negative and are multiplied.

Therefore , it equals - $\sqrt{ab}$.

For the OP.

score -1 · Answer 2 · edited Dec 19 '20 at 12:45

-1

$i = \sqrt{-1~}$ so $i^2 = -1$ and $i^3 = i^2 \cdot i = -1 \cdot \sqrt{-1~} = -\sqrt{-1~}$ Hope This helps!

edited Dec 19 '20 at 12:45

luxerhia

answered Dec 19 '20 at 09:11

Isaac

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Dec 19 '20 at 09:19
You are not wrong but you must use Math jax so that it easier for other users to understand your answer.@Issac – Dec 19 '20 at 10:07

2 Answers2