Confused with imaginary numbers

Question

In 9th grade I had an argument with my teacher that

${i}^{3}=i$

where $i=\sqrt{-1}$

But my teacher insisted (as is the accepted case) that:

${i}^{3}=-i$

My Solution:

${i}^3=(\sqrt{-1})^3$

${i}^3=\sqrt{(-1)^3}$

${i}^3=\sqrt{-1\times-1\times-1}$

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

${i}^3=i$

Generally accepted solution:

${i}^3=(\sqrt{-1})^3$

${i}^3=\sqrt{-1}\times\sqrt{-1}\times\sqrt{-1}$

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

${i}^3=-i$

What is so wrong with my approach? Is it not logical?

I am using the positive square root. There seems to be something about the order in which the power should be raised? There must be a logical reason, and I need help understanding it.

You might want to look at this: Simple complex Number Problem: 1=-1. — gebruiker, Apr 30 '15 at 21:29
As Avid's comment suggests, the issue is that you can't go from line 1 to 2. — Eric Stucky, Apr 30 '15 at 21:30
Of course $-1=1$ because $-1 = \sqrt{-1}^2 = \sqrt{(-1)^2} = \sqrt{1} = 1$. Thus $-\mathrm i = -1\mathrm i = 1\mathrm i = \mathrm i$. :-) — celtschk, Apr 30 '15 at 21:39
About "I am using the positive square root": No, the notion of "positive" is not defined for complex numbers. — Andreas Blass, Apr 30 '15 at 22:12
The problem is that there are two square roots of $-1$, and $i$ is just one of those. ${}\qquad{}$ — Michael Hardy, Apr 30 '15 at 22:14

score 13 · Answer 1 · answered Apr 30 '15 at 21:43

13

In a nutshell the square root function is not single valued. So you cannot always say $\sqrt{ab}=\sqrt a\cdot \sqrt b$. Otherwise you could also prove $1=-1$ as follows: $1=\sqrt1=\sqrt{-1\cdot-1}=\sqrt{-1}\sqrt{-1}=(\sqrt{-1})^2=-1$.

answered Apr 30 '15 at 21:43

Gregory Grant

14,874

score 12 · Answer 2 · answered Apr 30 '15 at 22:07

12

The error is in saying $i=\sqrt{-1}$. The correct statement is $i^2=-1$ and we're only allowed to use this property.

Thus $$ i^3=i^2\cdot i=(-1)\cdot i=-i $$

Never substitute $\sqrt{-1}$ for $i$: it just leads to errors, because the standard identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$ just holds for real and non negative numbers.

answered Apr 30 '15 at 22:07

egreg

238,574

I really like this answer +1 – imranfat May 01 '15 at 03:38

Christopher · Answer 3 · 2015-04-30T21:46:30.383

2

Here's a better proof than the given "generally accepted proof":

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

edited Apr 30 '15 at 21:46

answered Apr 30 '15 at 21:46

Christopher

7,137

score 2 · Answer 4 · answered Apr 30 '15 at 22:01

2

I think it is always healthier to avoid square roots. If we had equality $i^3=i$, then $$ -1=i\times i=i\times i^3=i^2\times i^2=(-1)(-1)=1. $$ Or, you could simply check: $i^3=i^2\times i=(-1)i=-i$.

answered Apr 30 '15 at 22:01

Martin Argerami

205,756

+1 Very true. I always introduce the imaginary unit as $i^2=-1$ and thus always avoid the square root! – imranfat May 01 '15 at 03:41

score 1 · Answer 5 · answered Apr 30 '15 at 21:41

1

The number $\sqrt{-1}$ is a two valued function with values $\{i,-i\}$ and hence the statement $$ \sqrt{-1}=i $$ is not true.

If you define the square function on the principal branch, then you can write $\sqrt{-1}=i$ but you can't pass the power into the square.

answered Apr 30 '15 at 21:41

k1.M

5,428

The idea to define $\sqrt{-1}=i$ is already an issue to begin with. See egreg's answer. – imranfat May 01 '15 at 03:40

Confused with imaginary numbers

5 Answers5