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 November 14th, 2013, 09:41 PM #1 Senior Member   Joined: Nov 2013 Posts: 160 Thanks: 7 what is the square root of +1 ? I have a question, $\sqrt{1}= +1$ $\sqrt{1}= -1$ because squaring on both sides gives the identity 1=1 . However, if you use imaginary number i which is defined by $\ i= sqrt{-1}$ you will get $1= \sqrt{1} = \sqrt{(-1)*(-1)} = \sqrt{(-1)} * \sqrt{(-1)} = i * i = i^2 = -1$ Because by definition of i $\ i^2= -1$ We will arrive at the result +1 = -1 , is there perhaps an error somewhere?
November 14th, 2013, 10:45 PM   #2
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Re: what is the square root of +1 ?

Quote:
 Originally Posted by TwoTwo I have a question, $\sqrt{1}= +1$ $\sqrt{1}= -1$
Actually, by definition:

$\sqrt{1}= +1$

That's why you have to write:

$\pm \sqrt{1}$

Quote:
 Originally Posted by TwoTwo However, if you use imaginary number i which is defined by $\ i= sqrt{-1}$
This is wrong, and you've shown exactly why it's wrong. i is defined as:

$i^2= 1$

So:

$sqrt{-1} \= \ \pm i$

With complex numbers there is no attempt to define square root (or the nth root) as any particular root. Unlike real numbers, there is no clear concept of the "positive one". So, for example:

$sqrt{i} \= \ \pm \frac{1}{2}(1+i)$

And the nth root will always be n complex numbers.

 November 15th, 2013, 12:47 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,933 Thanks: 2207 That's not right. By convention, ?(-1) = i, not -i. By convention, if z is a complex number with a non-zero imaginary part, ?z has an imaginary part with the same sign as the imaginary part of z. The equation (?xy) = (?x)(?y) holds for all positive values of x and y. As already shown, it doesn't hold for x = y = -1.
 November 15th, 2013, 02:19 AM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: what is the square root of +1 ? I must admit that's news to me. To take the argument set out by the OP: $1= \sqrt{1} = \sqrt{(-1)*(-1)} = \sqrt{(-1)} * \sqrt{(-1)} = i * i = i^2 = -1$ This must break down somewhere. I guess you have to say that, with that convention, for complex numbers: $\sqrt{zw} \neq \sqrt{z}\sqrt{w}$ ?
November 15th, 2013, 03:56 AM   #5
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Re: what is the square root of +1 ?

Quote:
 Originally Posted by Pero I must admit that's news to me. To take the argument set out by the OP: $1= \sqrt{1} = \sqrt{(-1)*(-1)} = \sqrt{(-1)} * \sqrt{(-1)} = i * i = i^2 = -1$ This must break down somewhere. I guess you have to say that, with that convention, for complex numbers: $\sqrt{zw} \neq \sqrt{z}\sqrt{w}$ ?

Look at my first post, 1 has two roots
$\sqrt{1}= +1$
$\sqrt{1}= -1$

so perhaps
$\sqrt{zw}= \sqrt{z}\sqrt{w}$
holds for negative values of z and w too.

Look at how Wolfram Alpha calculates square roots (also known as 2nd roots) of 1:
http://www.wolframalpha.com/input/?i=sqrt1
You can see two real roots +1 (real root, principal root) and -1 (real root)

 November 15th, 2013, 05:04 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,933 Thanks: 2207 Sometimes it works for complex numbers and sometimes it doesn't.
 November 15th, 2013, 10:14 PM #7 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: what is the square root of +1 ? $1 \= \ \sqrt{1} \ = \ \sqrt{(-1) \cdot (-1) \cdot (-1) \cdot (-1)} \ = \sqrt{(-1)} \cdot \sqrt{(-1)} \cdot \sqrt{(-1)} \cdot \sqrt{(-1)} \ = \ i \cdot i \cdot i \cdot i \ = \ i^2 \cdot i^2 \ = (-1) \cdot (-1) \ = \ 1$ So everything is right as rain again.
November 16th, 2013, 02:21 AM   #8
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Re: what is the square root of +1 ?

Quote:
 Originally Posted by agentredlum $1 \= \ \sqrt{1} \ = \ \sqrt{(-1) \cdot (-1) \cdot (-1) \cdot (-1)} \ = \sqrt{(-1)} \cdot \sqrt{(-1)} \cdot \sqrt{(-1)} \cdot \sqrt{(-1)} \ = \ i \cdot i \cdot i \cdot i \ = \ i^2 \cdot i^2 \ = (-1) \cdot (-1) \ = \ 1$ So everything is right as rain again.

Ok, very good. Consider now

$\ i \cdot i \cdot i \cdot i \= \ i^2 \cdot i^2 \ = \ i^4 \ = 1$

but what is
$\sqrt{(\ i^4 \)} ?$

Is $\sqrt{(\ i^4 \)}= \ i^2 \ = -1$

Or is $\sqrt{(\ i^4 \)}= \ sqrt{(-1)\cdot (-1)\cdot(-1)\cdot(-1)} \ = +1 ?$

 November 16th, 2013, 05:07 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,933 Thanks: 2207 $\sqrt{ i^{\small4}}\,=\,sqrt1\,=\,1$ The rule that (z^m)^n = z^(mn) always works if z > 0 and both m and n are real, but only sometimes works otherwise.
November 16th, 2013, 10:59 AM   #10
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Re:

Quote:
 Originally Posted by skipjack Sometimes it works for complex numbers and sometimes it doesn't.
This is a strange situation, and I try to understand what is the problem.
Again I write the same formulas but now a little differently. How to prove that
$\sqrt{z}\sqrt{w}=\sqrt{zw}$
holds even when both z and w are negative:

$i^2= i*i = \sqrt{(-1)} * \sqrt{(-1)} = \sqrt{(-1)*(-1)} = \sqrt{(+1)*(+1)} = \sqrt{1} = -1$

Because
$\sqrt{1}= -1$

And this time the paradox +1 = -1 is hidden.

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