Solutions of $z=dfracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i$Locating possible complex solutions to an equation involving a square rootFinding the square root of a complex number - why two solutions instead of four?square root of $frac2+sqrt34$$(n-1)^textth$ root of conjugateWhy don't $sqrta*b=sqrta*sqrtb$ holds true when one of $a$ & $b$ is imaginary number?Solve $x^4 + 64 = 0$Roots of complex quadratic polynomialWhy $Releft(e^sqrt-log xright)=Imleft(e^sqrt-log xright)$ has no real solutions?Are there multiple solutions to a fractional root/radical (over the complex numbers)?Question about this exponent problem.
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Solutions of $z=dfracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i$
Locating possible complex solutions to an equation involving a square rootFinding the square root of a complex number - why two solutions instead of four?square root of $frac2+sqrt34$$(n-1)^textth$ root of conjugateWhy don't $sqrta*b=sqrta*sqrtb$ holds true when one of $a$ & $b$ is imaginary number?Solve $x^4 + 64 = 0$Roots of complex quadratic polynomialWhy $Releft(e^sqrt-log xright)=Imleft(e^sqrt-log xright)$ has no real solutions?Are there multiple solutions to a fractional root/radical (over the complex numbers)?Question about this exponent problem.
$begingroup$
Prove that the solution of $z=dfracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i$ are $dfrac5i2$ and $dfrac-2i5$
I know that If I write $sqrt-21+20i=pm(2+5i)$ and $sqrt-21-20i=pm(2-5i)$, then I will get the required solutions. But, when I do the following :
beginalign
z&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i\
&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i.fracsqrt-21+20i-sqrt-21-20isqrt-21+20i-sqrt-21-20i\
&=frac(sqrt-21+20i-sqrt-21-20i)^2(sqrt-21+20i)^2-(sqrt-21-20i)^2\
&=frac-21+20i-21-20i-2sqrt21^2+20^2=841-21+20i+21+20i\
&=frac-21-sqrt84120i=frac-21-2920i=frac5i2
endalign
Why am I missing the solution $dfrac-2i5$ in my attempt ?
If I were to take $sqrt841=-29$ I think I'll get the other solution but, I always considered $sqrtx$ as the +ve square root and $-sqrtx$ the -ve square root of the number $x$.
complex-numbers
asked Mar 16 at 21:27
ss1729ss1729
2,04411124
$endgroup$
add a comment |
$begingroup$
Prove that the solution of $z=dfracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i$ are $dfrac5i2$ and $dfrac-2i5$
I know that If I write $sqrt-21+20i=pm(2+5i)$ and $sqrt-21-20i=pm(2-5i)$, then I will get the required solutions. But, when I do the following :
beginalign
z&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i\
&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i.fracsqrt-21+20i-sqrt-21-20isqrt-21+20i-sqrt-21-20i\
&=frac(sqrt-21+20i-sqrt-21-20i)^2(sqrt-21+20i)^2-(sqrt-21-20i)^2\
&=frac-21+20i-21-20i-2sqrt21^2+20^2=841-21+20i+21+20i\
&=frac-21-sqrt84120i=frac-21-2920i=frac5i2
endalign
Why am I missing the solution $dfrac-2i5$ in my attempt ?
If I were to take $sqrt841=-29$ I think I'll get the other solution but, I always considered $sqrtx$ as the +ve square root and $-sqrtx$ the -ve square root of the number $x$.
complex-numbers
asked Mar 16 at 21:27
ss1729ss1729
2,04411124
$endgroup$
add a comment |
$begingroup$
Prove that the solution of $z=dfracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i$ are $dfrac5i2$ and $dfrac-2i5$
I know that If I write $sqrt-21+20i=pm(2+5i)$ and $sqrt-21-20i=pm(2-5i)$, then I will get the required solutions. But, when I do the following :
beginalign
z&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i\
&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i.fracsqrt-21+20i-sqrt-21-20isqrt-21+20i-sqrt-21-20i\
&=frac(sqrt-21+20i-sqrt-21-20i)^2(sqrt-21+20i)^2-(sqrt-21-20i)^2\
&=frac-21+20i-21-20i-2sqrt21^2+20^2=841-21+20i+21+20i\
&=frac-21-sqrt84120i=frac-21-2920i=frac5i2
endalign
Why am I missing the solution $dfrac-2i5$ in my attempt ?
If I were to take $sqrt841=-29$ I think I'll get the other solution but, I always considered $sqrtx$ as the +ve square root and $-sqrtx$ the -ve square root of the number $x$.
complex-numbers
asked Mar 16 at 21:27
ss1729ss1729
2,04411124
$endgroup$
Prove that the solution of $z=dfracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i$ are $dfrac5i2$ and $dfrac-2i5$
I know that If I write $sqrt-21+20i=pm(2+5i)$ and $sqrt-21-20i=pm(2-5i)$, then I will get the required solutions. But, when I do the following :
beginalign
z&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i\
&=fracsqrt-21+20i-sqrt-21-20isqrt-21+20i+sqrt-21-20i.fracsqrt-21+20i-sqrt-21-20isqrt-21+20i-sqrt-21-20i\
&=frac(sqrt-21+20i-sqrt-21-20i)^2(sqrt-21+20i)^2-(sqrt-21-20i)^2\
&=frac-21+20i-21-20i-2sqrt21^2+20^2=841-21+20i+21+20i\
&=frac-21-sqrt84120i=frac-21-2920i=frac5i2
endalign
Why am I missing the solution $dfrac-2i5$ in my attempt ?
If I were to take $sqrt841=-29$ I think I'll get the other solution but, I always considered $sqrtx$ as the +ve square root and $-sqrtx$ the -ve square root of the number $x$.
complex-numbers
complex-numbers
asked Mar 16 at 21:27
ss1729ss1729
2,04411124
asked Mar 16 at 21:27
ss1729ss1729
2,04411124
edited Mar 16 at 21:53
ss1729
asked Mar 16 at 21:27
ss1729ss1729
2,04411124
asked Mar 16 at 21:27
ss1729ss1729
2,04411124
asked Mar 16 at 21:27
ss1729ss1729
2,04411124
2,04411124
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Remember that $841$ has two square roots.
answered Mar 16 at 21:29
ArthurArthur
120k7119203
$endgroup$
$begingroup$
thats right. but usually mean $sqrtx$, +ve square root of $x$ and $-sqrtx$, -ve square root of $x$ right ?
$endgroup$
– ss1729
Mar 16 at 21:45
$begingroup$
But here, we don't. Otherwise the expression on the right has exactly one value.
$endgroup$
– enedil
Mar 16 at 21:49
$begingroup$
@enedil i agree if I consider that I'll get the other solution. But, If I wee to twist the convention here I need to do the same for any problem right ?. I think I am particularly confused about the fact that $sqrtx$ only mean the +ve square root usually.
$endgroup$
– ss1729
Mar 16 at 22:00
$begingroup$
@ss1729 But in this specific case, $sqrt841$ actually appears as the product of two complex square roots ($sqrt-21+20icdotsqrt-21-20i$), and I don't think you have any issues accepting that those are multivalued.
$endgroup$
– Arthur
Mar 16 at 22:07
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Remember that $841$ has two square roots.
answered Mar 16 at 21:29
ArthurArthur
120k7119203
$endgroup$
$begingroup$
thats right. but usually mean $sqrtx$, +ve square root of $x$ and $-sqrtx$, -ve square root of $x$ right ?
$endgroup$
– ss1729
Mar 16 at 21:45
$begingroup$
But here, we don't. Otherwise the expression on the right has exactly one value.
$endgroup$
– enedil
Mar 16 at 21:49
$begingroup$
@enedil i agree if I consider that I'll get the other solution. But, If I wee to twist the convention here I need to do the same for any problem right ?. I think I am particularly confused about the fact that $sqrtx$ only mean the +ve square root usually.
$endgroup$
– ss1729
Mar 16 at 22:00
$begingroup$
@ss1729 But in this specific case, $sqrt841$ actually appears as the product of two complex square roots ($sqrt-21+20icdotsqrt-21-20i$), and I don't think you have any issues accepting that those are multivalued.
$endgroup$
– Arthur
Mar 16 at 22:07
add a comment |
$begingroup$
Remember that $841$ has two square roots.
answered Mar 16 at 21:29
ArthurArthur
120k7119203
$endgroup$
$begingroup$
thats right. but usually mean $sqrtx$, +ve square root of $x$ and $-sqrtx$, -ve square root of $x$ right ?
$endgroup$
– ss1729
Mar 16 at 21:45
$begingroup$
But here, we don't. Otherwise the expression on the right has exactly one value.
$endgroup$
– enedil
Mar 16 at 21:49
$begingroup$
@enedil i agree if I consider that I'll get the other solution. But, If I wee to twist the convention here I need to do the same for any problem right ?. I think I am particularly confused about the fact that $sqrtx$ only mean the +ve square root usually.
$endgroup$
– ss1729
Mar 16 at 22:00
$begingroup$
@ss1729 But in this specific case, $sqrt841$ actually appears as the product of two complex square roots ($sqrt-21+20icdotsqrt-21-20i$), and I don't think you have any issues accepting that those are multivalued.
$endgroup$
– Arthur
Mar 16 at 22:07
add a comment |
$begingroup$
Remember that $841$ has two square roots.
answered Mar 16 at 21:29
ArthurArthur
120k7119203
$endgroup$
Remember that $841$ has two square roots.
answered Mar 16 at 21:29
ArthurArthur
120k7119203
answered Mar 16 at 21:29
ArthurArthur
120k7119203
answered Mar 16 at 21:29
ArthurArthur
120k7119203
answered Mar 16 at 21:29
ArthurArthur
120k7119203
120k7119203
$begingroup$
thats right. but usually mean $sqrtx$, +ve square root of $x$ and $-sqrtx$, -ve square root of $x$ right ?
$endgroup$
– ss1729
Mar 16 at 21:45
$begingroup$
But here, we don't. Otherwise the expression on the right has exactly one value.
$endgroup$
– enedil
Mar 16 at 21:49
$begingroup$
@enedil i agree if I consider that I'll get the other solution. But, If I wee to twist the convention here I need to do the same for any problem right ?. I think I am particularly confused about the fact that $sqrtx$ only mean the +ve square root usually.
$endgroup$
– ss1729
Mar 16 at 22:00
$begingroup$
@ss1729 But in this specific case, $sqrt841$ actually appears as the product of two complex square roots ($sqrt-21+20icdotsqrt-21-20i$), and I don't think you have any issues accepting that those are multivalued.
$endgroup$
– Arthur
Mar 16 at 22:07
add a comment |
$begingroup$
thats right. but usually mean $sqrtx$, +ve square root of $x$ and $-sqrtx$, -ve square root of $x$ right ?
$endgroup$
– ss1729
Mar 16 at 21:45
$begingroup$
But here, we don't. Otherwise the expression on the right has exactly one value.
$endgroup$
– enedil
Mar 16 at 21:49
$begingroup$
@enedil i agree if I consider that I'll get the other solution. But, If I wee to twist the convention here I need to do the same for any problem right ?. I think I am particularly confused about the fact that $sqrtx$ only mean the +ve square root usually.
$endgroup$
– ss1729
Mar 16 at 22:00
$begingroup$
@ss1729 But in this specific case, $sqrt841$ actually appears as the product of two complex square roots ($sqrt-21+20icdotsqrt-21-20i$), and I don't think you have any issues accepting that those are multivalued.
$endgroup$
– Arthur
Mar 16 at 22:07
$begingroup$
thats right. but usually mean $sqrtx$, +ve square root of $x$ and $-sqrtx$, -ve square root of $x$ right ?
$endgroup$
– ss1729
Mar 16 at 21:45
$begingroup$
thats right. but usually mean $sqrtx$, +ve square root of $x$ and $-sqrtx$, -ve square root of $x$ right ?
$endgroup$
– ss1729
Mar 16 at 21:45
$begingroup$
But here, we don't. Otherwise the expression on the right has exactly one value.
$endgroup$
– enedil
Mar 16 at 21:49
$begingroup$
But here, we don't. Otherwise the expression on the right has exactly one value.
$endgroup$
– enedil
Mar 16 at 21:49
$begingroup$
@enedil i agree if I consider that I'll get the other solution. But, If I wee to twist the convention here I need to do the same for any problem right ?. I think I am particularly confused about the fact that $sqrtx$ only mean the +ve square root usually.
$endgroup$
– ss1729
Mar 16 at 22:00
$begingroup$
@enedil i agree if I consider that I'll get the other solution. But, If I wee to twist the convention here I need to do the same for any problem right ?. I think I am particularly confused about the fact that $sqrtx$ only mean the +ve square root usually.
$endgroup$
– ss1729
Mar 16 at 22:00
$begingroup$
@ss1729 But in this specific case, $sqrt841$ actually appears as the product of two complex square roots ($sqrt-21+20icdotsqrt-21-20i$), and I don't think you have any issues accepting that those are multivalued.
$endgroup$
– Arthur
Mar 16 at 22:07
$begingroup$
@ss1729 But in this specific case, $sqrt841$ actually appears as the product of two complex square roots ($sqrt-21+20icdotsqrt-21-20i$), and I don't think you have any issues accepting that those are multivalued.
$endgroup$
– Arthur
Mar 16 at 22:07
add a comment |
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