Mixing
Is $sqrt{-i sqrt{5}}=-isqrt[4]{-5}$?
$begingroup$
I have tried doing
$sqrt{-i sqrt{5}}$
$=sqrt{-sqrt{-1} sqrt{5}}$
$=sqrt{-sqrt{-5}}$
$=isqrt{sqrt{-5}}$
$=isqrt[4]{-5}$
But Wolfram says that the answer is $-isqrt[4]{-5}$.
Where is my error?
complex-numbers
edited Jan 30 at 13:42
J. W. Tanner
4,4401320
asked Jan 30 at 12:27
retro_varretro_var
625215
$endgroup$
add a comment |
$begingroup$
I have tried doing
$sqrt{-i sqrt{5}}$
$=sqrt{-sqrt{-1} sqrt{5}}$
$=sqrt{-sqrt{-5}}$
$=isqrt{sqrt{-5}}$
$=isqrt[4]{-5}$
But Wolfram says that the answer is $-isqrt[4]{-5}$.
Where is my error?
complex-numbers
edited Jan 30 at 13:42
J. W. Tanner
4,4401320
asked Jan 30 at 12:27
retro_varretro_var
625215
$endgroup$
4
$begingroup$
$sqrt{-sqrt{-5}}≠isqrt{sqrt{-5}}$
$endgroup$
– Holo
Jan 30 at 12:32
1
$begingroup$
$-isqrt{5}$ has two square roots. One is the negative of the other, as you found.
$endgroup$
– GEdgar
Jan 30 at 13:29
add a comment |
$begingroup$
I have tried doing
$sqrt{-i sqrt{5}}$
$=sqrt{-sqrt{-1} sqrt{5}}$
$=sqrt{-sqrt{-5}}$
$=isqrt{sqrt{-5}}$
$=isqrt[4]{-5}$
But Wolfram says that the answer is $-isqrt[4]{-5}$.
Where is my error?
complex-numbers
edited Jan 30 at 13:42
J. W. Tanner
4,4401320
asked Jan 30 at 12:27
retro_varretro_var
625215
$endgroup$
I have tried doing
$sqrt{-i sqrt{5}}$
$=sqrt{-sqrt{-1} sqrt{5}}$
$=sqrt{-sqrt{-5}}$
$=isqrt{sqrt{-5}}$
$=isqrt[4]{-5}$
But Wolfram says that the answer is $-isqrt[4]{-5}$.
Where is my error?
complex-numbers
complex-numbers
edited Jan 30 at 13:42
J. W. Tanner
4,4401320
asked Jan 30 at 12:27
retro_varretro_var
625215
edited Jan 30 at 13:42
J. W. Tanner
4,4401320
asked Jan 30 at 12:27
retro_varretro_var
625215
edited Jan 30 at 13:42
J. W. Tanner
4,4401320
edited Jan 30 at 13:42
J. W. Tanner
4,4401320
edited Jan 30 at 13:42
J. W. Tanner
4,4401320
4,4401320
asked Jan 30 at 12:27
retro_varretro_var
625215
asked Jan 30 at 12:27
retro_varretro_var
625215
asked Jan 30 at 12:27
retro_varretro_var
625215
625215
4
$begingroup$
$sqrt{-sqrt{-5}}≠isqrt{sqrt{-5}}$
$endgroup$
– Holo
Jan 30 at 12:32
1
$begingroup$
$-isqrt{5}$ has two square roots. One is the negative of the other, as you found.
$endgroup$
– GEdgar
Jan 30 at 13:29
add a comment |
4
$begingroup$
$sqrt{-sqrt{-5}}≠isqrt{sqrt{-5}}$
$endgroup$
– Holo
Jan 30 at 12:32
1
$begingroup$
$-isqrt{5}$ has two square roots. One is the negative of the other, as you found.
$endgroup$
– GEdgar
Jan 30 at 13:29
4
4
$begingroup$
$sqrt{-sqrt{-5}}≠isqrt{sqrt{-5}}$
$endgroup$
– Holo
Jan 30 at 12:32
$begingroup$
$sqrt{-sqrt{-5}}≠isqrt{sqrt{-5}}$
$endgroup$
– Holo
Jan 30 at 12:32
1
1
$begingroup$
$-isqrt{5}$ has two square roots. One is the negative of the other, as you found.
$endgroup$
– GEdgar
Jan 30 at 13:29
$begingroup$
$-isqrt{5}$ has two square roots. One is the negative of the other, as you found.
$endgroup$
– GEdgar
Jan 30 at 13:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It depends on the definition of square root over the field of complex numbers. Since equation $x^n=c$ has $n$ roots for any $cneq 0$, we can either define n-th root as a multivalue function (and thus your answer is just the other value of this function, that was found by WA) or we can understand n-th root as principal root (restrict the answer to have argument in half-interval $(-pi/n,pi/n]$), in that case $sqrt{a}sqrt{b}neqsqrt{ab}$ and that's your mistake.
answered Jan 30 at 13:00
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
It depends on the definition of square root over the field of complex numbers. Since equation $x^n=c$ has $n$ roots for any $cneq 0$, we can either define n-th root as a multivalue function (and thus your answer is just the other value of this function, that was found by WA) or we can understand n-th root as principal root (restrict the answer to have argument in half-interval $(-pi/n,pi/n]$), in that case $sqrt{a}sqrt{b}neqsqrt{ab}$ and that's your mistake.
answered Jan 30 at 13:00
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
$begingroup$
It depends on the definition of square root over the field of complex numbers. Since equation $x^n=c$ has $n$ roots for any $cneq 0$, we can either define n-th root as a multivalue function (and thus your answer is just the other value of this function, that was found by WA) or we can understand n-th root as principal root (restrict the answer to have argument in half-interval $(-pi/n,pi/n]$), in that case $sqrt{a}sqrt{b}neqsqrt{ab}$ and that's your mistake.
answered Jan 30 at 13:00
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
$begingroup$
It depends on the definition of square root over the field of complex numbers. Since equation $x^n=c$ has $n$ roots for any $cneq 0$, we can either define n-th root as a multivalue function (and thus your answer is just the other value of this function, that was found by WA) or we can understand n-th root as principal root (restrict the answer to have argument in half-interval $(-pi/n,pi/n]$), in that case $sqrt{a}sqrt{b}neqsqrt{ab}$ and that's your mistake.
answered Jan 30 at 13:00
Vasily MitchVasily Mitch
2,6791312
$endgroup$
It depends on the definition of square root over the field of complex numbers. Since equation $x^n=c$ has $n$ roots for any $cneq 0$, we can either define n-th root as a multivalue function (and thus your answer is just the other value of this function, that was found by WA) or we can understand n-th root as principal root (restrict the answer to have argument in half-interval $(-pi/n,pi/n]$), in that case $sqrt{a}sqrt{b}neqsqrt{ab}$ and that's your mistake.
answered Jan 30 at 13:00
Vasily MitchVasily Mitch
2,6791312
answered Jan 30 at 13:00
Vasily MitchVasily Mitch
2,6791312
answered Jan 30 at 13:00
Vasily MitchVasily Mitch
2,6791312
answered Jan 30 at 13:00
Vasily MitchVasily Mitch
2,6791312
2,6791312
add a comment |
add a comment |
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4
$begingroup$
$sqrt{-sqrt{-5}}≠isqrt{sqrt{-5}}$
$endgroup$
– Holo
Jan 30 at 12:32
1
$begingroup$
$-isqrt{5}$ has two square roots. One is the negative of the other, as you found.
$endgroup$
– GEdgar
Jan 30 at 13:29