Problem in complex number multiplication [duplicate]
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This question already has an answer here:
Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?
9 answers
I want to know $sqrt{-m}sqrt{-n}=$? I tried in the following ways:
Way 1:$$sqrt{-m}sqrt{-n}=sqrt{(-m)(-n)}=sqrt{mn}.$$
Way 2:$$sqrt{-m}sqrt{-n}=sqrt{m}isqrt{n}i=sqrt{mn}i^2=-sqrt{mn}$$
So I got two different values for the same $sqrt{-m}sqrt{-n}$. How can that be possible?
complex-numbers complex-multiplication
edited Dec 1 '18 at 8:18
Saad
19.7k92352
asked Dec 1 '18 at 8:02
Asif IqubalAsif Iqubal
1397
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marked as duplicate by Saad, Lord Shark the Unknown, Watson, Cesareo, Brahadeesh Dec 1 '18 at 10:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?
9 answers
I want to know $sqrt{-m}sqrt{-n}=$? I tried in the following ways:
Way 1:$$sqrt{-m}sqrt{-n}=sqrt{(-m)(-n)}=sqrt{mn}.$$
Way 2:$$sqrt{-m}sqrt{-n}=sqrt{m}isqrt{n}i=sqrt{mn}i^2=-sqrt{mn}$$
So I got two different values for the same $sqrt{-m}sqrt{-n}$. How can that be possible?
complex-numbers complex-multiplication
edited Dec 1 '18 at 8:18
Saad
19.7k92352
asked Dec 1 '18 at 8:02
Asif IqubalAsif Iqubal
1397
$endgroup$
marked as duplicate by Saad, Lord Shark the Unknown, Watson, Cesareo, Brahadeesh Dec 1 '18 at 10:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?
9 answers
I want to know $sqrt{-m}sqrt{-n}=$? I tried in the following ways:
Way 1:$$sqrt{-m}sqrt{-n}=sqrt{(-m)(-n)}=sqrt{mn}.$$
Way 2:$$sqrt{-m}sqrt{-n}=sqrt{m}isqrt{n}i=sqrt{mn}i^2=-sqrt{mn}$$
So I got two different values for the same $sqrt{-m}sqrt{-n}$. How can that be possible?
complex-numbers complex-multiplication
edited Dec 1 '18 at 8:18
Saad
19.7k92352
asked Dec 1 '18 at 8:02
Asif IqubalAsif Iqubal
1397
$endgroup$
This question already has an answer here:
Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?
9 answers
I want to know $sqrt{-m}sqrt{-n}=$? I tried in the following ways:
Way 1:$$sqrt{-m}sqrt{-n}=sqrt{(-m)(-n)}=sqrt{mn}.$$
Way 2:$$sqrt{-m}sqrt{-n}=sqrt{m}isqrt{n}i=sqrt{mn}i^2=-sqrt{mn}$$
So I got two different values for the same $sqrt{-m}sqrt{-n}$. How can that be possible?
This question already has an answer here:
Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?
9 answers
complex-numbers complex-multiplication
complex-numbers complex-multiplication
edited Dec 1 '18 at 8:18
Saad
19.7k92352
asked Dec 1 '18 at 8:02
Asif IqubalAsif Iqubal
1397
edited Dec 1 '18 at 8:18
Saad
19.7k92352
asked Dec 1 '18 at 8:02
Asif IqubalAsif Iqubal
1397
edited Dec 1 '18 at 8:18
Saad
19.7k92352
edited Dec 1 '18 at 8:18
Saad
19.7k92352
edited Dec 1 '18 at 8:18
Saad
19.7k92352
19.7k92352
asked Dec 1 '18 at 8:02
Asif IqubalAsif Iqubal
1397
asked Dec 1 '18 at 8:02
Asif IqubalAsif Iqubal
1397
asked Dec 1 '18 at 8:02
Asif IqubalAsif Iqubal
1397
1397
marked as duplicate by Saad, Lord Shark the Unknown, Watson, Cesareo, Brahadeesh Dec 1 '18 at 10:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Saad, Lord Shark the Unknown, Watson, Cesareo, Brahadeesh Dec 1 '18 at 10:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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The rule $sqrt{a} cdot sqrt{b} = sqrt{a cdot b}$ doesn't necessarily apply when $a < 0$ or $b < 0$. For example, $$-1 = icdot i = sqrt{-1} cdot sqrt{-1} neq sqrt{(-1)cdot(-1)} = sqrt{1} = 1$$
Therefore, the 1st way is incorrect.
answered Dec 1 '18 at 8:40
AlephZeroAlephZero
22616
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The rule $sqrt{a} cdot sqrt{b} = sqrt{a cdot b}$ doesn't necessarily apply when $a < 0$ or $b < 0$. For example, $$-1 = icdot i = sqrt{-1} cdot sqrt{-1} neq sqrt{(-1)cdot(-1)} = sqrt{1} = 1$$
Therefore, the 1st way is incorrect.
answered Dec 1 '18 at 8:40
AlephZeroAlephZero
22616
$endgroup$
add a comment |
$begingroup$
The rule $sqrt{a} cdot sqrt{b} = sqrt{a cdot b}$ doesn't necessarily apply when $a < 0$ or $b < 0$. For example, $$-1 = icdot i = sqrt{-1} cdot sqrt{-1} neq sqrt{(-1)cdot(-1)} = sqrt{1} = 1$$
Therefore, the 1st way is incorrect.
answered Dec 1 '18 at 8:40
AlephZeroAlephZero
22616
$endgroup$
add a comment |
$begingroup$
The rule $sqrt{a} cdot sqrt{b} = sqrt{a cdot b}$ doesn't necessarily apply when $a < 0$ or $b < 0$. For example, $$-1 = icdot i = sqrt{-1} cdot sqrt{-1} neq sqrt{(-1)cdot(-1)} = sqrt{1} = 1$$
Therefore, the 1st way is incorrect.
answered Dec 1 '18 at 8:40
AlephZeroAlephZero
22616
$endgroup$
The rule $sqrt{a} cdot sqrt{b} = sqrt{a cdot b}$ doesn't necessarily apply when $a < 0$ or $b < 0$. For example, $$-1 = icdot i = sqrt{-1} cdot sqrt{-1} neq sqrt{(-1)cdot(-1)} = sqrt{1} = 1$$
Therefore, the 1st way is incorrect.
answered Dec 1 '18 at 8:40
AlephZeroAlephZero
22616
answered Dec 1 '18 at 8:40
AlephZeroAlephZero
22616
answered Dec 1 '18 at 8:40
AlephZeroAlephZero
22616
answered Dec 1 '18 at 8:40
AlephZeroAlephZero
22616
22616
add a comment |
add a comment |
