Find real part of messy complex fraction?
$begingroup$
I have this messy complex fraction, where the LHS is real. But how can I find the real part of the RHS?
$$
frac{R277-X115+i(R115+X277)}{R+277+i(115+X)}=300,
$$
where $R, Xin mathbb R$.
complex-numbers
asked Nov 13 '18 at 10:36
JDoeDoeJDoeDoe
7851616
$endgroup$
add a comment |
$begingroup$
I have this messy complex fraction, where the LHS is real. But how can I find the real part of the RHS?
$$
frac{R277-X115+i(R115+X277)}{R+277+i(115+X)}=300,
$$
where $R, Xin mathbb R$.
complex-numbers
asked Nov 13 '18 at 10:36
JDoeDoeJDoeDoe
7851616
$endgroup$
$begingroup$
Well, $Re(300)=300$
$endgroup$
– gammatester
Nov 13 '18 at 10:39
1
$begingroup$
Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
$endgroup$
– saulspatz
Nov 13 '18 at 10:43
add a comment |
$begingroup$
I have this messy complex fraction, where the LHS is real. But how can I find the real part of the RHS?
$$
frac{R277-X115+i(R115+X277)}{R+277+i(115+X)}=300,
$$
where $R, Xin mathbb R$.
complex-numbers
asked Nov 13 '18 at 10:36
JDoeDoeJDoeDoe
7851616
$endgroup$
I have this messy complex fraction, where the LHS is real. But how can I find the real part of the RHS?
$$
frac{R277-X115+i(R115+X277)}{R+277+i(115+X)}=300,
$$
where $R, Xin mathbb R$.
complex-numbers
complex-numbers
asked Nov 13 '18 at 10:36
JDoeDoeJDoeDoe
7851616
asked Nov 13 '18 at 10:36
JDoeDoeJDoeDoe
7851616
asked Nov 13 '18 at 10:36
JDoeDoeJDoeDoe
7851616
asked Nov 13 '18 at 10:36
JDoeDoeJDoeDoe
7851616
asked Nov 13 '18 at 10:36
JDoeDoeJDoeDoe
7851616
7851616
$begingroup$
Well, $Re(300)=300$
$endgroup$
– gammatester
Nov 13 '18 at 10:39
1
$begingroup$
Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
$endgroup$
– saulspatz
Nov 13 '18 at 10:43
add a comment |
$begingroup$
Well, $Re(300)=300$
$endgroup$
– gammatester
Nov 13 '18 at 10:39
1
$begingroup$
Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
$endgroup$
– saulspatz
Nov 13 '18 at 10:43
$begingroup$
Well, $Re(300)=300$
$endgroup$
– gammatester
Nov 13 '18 at 10:39
$begingroup$
Well, $Re(300)=300$
$endgroup$
– gammatester
Nov 13 '18 at 10:39
1
1
$begingroup$
Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
$endgroup$
– saulspatz
Nov 13 '18 at 10:43
$begingroup$
Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
$endgroup$
– saulspatz
Nov 13 '18 at 10:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I believe that R=resistance and X=impedance.
So:
$R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$
begin{cases}
R277−X115 = 300 (R+277) \
R115+X277 = 300 (115+X) \
end{cases}
begin{cases}
R277−X115 = 300R+300*277 \
R115+X277 = 300*115+300X \
end{cases}
begin{cases}
R277-300R-X115 = 300*277 \
R115+X277-300X = 300*115 \
end{cases}
begin{cases}
-23R-115X = 300*277 = 83100 \
115R+577X = 300*115 = 34500 \
end{cases}
begin{cases}
R=-1128613 \
X= 225000 \
end{cases}
But R is negative! Can anybody verify ?
answered Nov 22 '18 at 21:06
cgiovanardicgiovanardi
764411
$endgroup$
add a comment |
$begingroup$
I'm assuming that you flipped right and left (since the real part of $300$ is clearly $300$). In general, to put a complex fraction in standard $x+yi$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:
$$ frac{a+bi}{c+di} ~~=~~ frac{a+bi}{c+di} cdot frac{c-di}{c-di} ~~=~~ frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$
Therefore, the real part of this fraction is
$$ frac{ac+bd}{c^2+d^2}. $$
In your case:
- $a=R277-X115$
- $b=R115+X277$
- $c=R+277$
- $d=X+115$
Plugging these into the formula above, we get
$$frac{[R^2277-RX115+R277^2-X(277cdot115)]+[RX115+X^2277+R115^2+X(277cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$
Cancelling terms in the numerator gives
$$=frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$
This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $115^2=13225$, and so on.
In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.
The resulting system is best solved by machine: here's the computation in WolframAlpha.
answered Dec 23 '18 at 4:57
aleph_twoaleph_two
27412
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe that R=resistance and X=impedance.
So:
$R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$
begin{cases}
R277−X115 = 300 (R+277) \
R115+X277 = 300 (115+X) \
end{cases}
begin{cases}
R277−X115 = 300R+300*277 \
R115+X277 = 300*115+300X \
end{cases}
begin{cases}
R277-300R-X115 = 300*277 \
R115+X277-300X = 300*115 \
end{cases}
begin{cases}
-23R-115X = 300*277 = 83100 \
115R+577X = 300*115 = 34500 \
end{cases}
begin{cases}
R=-1128613 \
X= 225000 \
end{cases}
But R is negative! Can anybody verify ?
answered Nov 22 '18 at 21:06
cgiovanardicgiovanardi
764411
$endgroup$
add a comment |
$begingroup$
I believe that R=resistance and X=impedance.
So:
$R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$
begin{cases}
R277−X115 = 300 (R+277) \
R115+X277 = 300 (115+X) \
end{cases}
begin{cases}
R277−X115 = 300R+300*277 \
R115+X277 = 300*115+300X \
end{cases}
begin{cases}
R277-300R-X115 = 300*277 \
R115+X277-300X = 300*115 \
end{cases}
begin{cases}
-23R-115X = 300*277 = 83100 \
115R+577X = 300*115 = 34500 \
end{cases}
begin{cases}
R=-1128613 \
X= 225000 \
end{cases}
But R is negative! Can anybody verify ?
answered Nov 22 '18 at 21:06
cgiovanardicgiovanardi
764411
$endgroup$
add a comment |
$begingroup$
I believe that R=resistance and X=impedance.
So:
$R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$
begin{cases}
R277−X115 = 300 (R+277) \
R115+X277 = 300 (115+X) \
end{cases}
begin{cases}
R277−X115 = 300R+300*277 \
R115+X277 = 300*115+300X \
end{cases}
begin{cases}
R277-300R-X115 = 300*277 \
R115+X277-300X = 300*115 \
end{cases}
begin{cases}
-23R-115X = 300*277 = 83100 \
115R+577X = 300*115 = 34500 \
end{cases}
begin{cases}
R=-1128613 \
X= 225000 \
end{cases}
But R is negative! Can anybody verify ?
answered Nov 22 '18 at 21:06
cgiovanardicgiovanardi
764411
$endgroup$
I believe that R=resistance and X=impedance.
So:
$R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$
begin{cases}
R277−X115 = 300 (R+277) \
R115+X277 = 300 (115+X) \
end{cases}
begin{cases}
R277−X115 = 300R+300*277 \
R115+X277 = 300*115+300X \
end{cases}
begin{cases}
R277-300R-X115 = 300*277 \
R115+X277-300X = 300*115 \
end{cases}
begin{cases}
-23R-115X = 300*277 = 83100 \
115R+577X = 300*115 = 34500 \
end{cases}
begin{cases}
R=-1128613 \
X= 225000 \
end{cases}
But R is negative! Can anybody verify ?
answered Nov 22 '18 at 21:06
cgiovanardicgiovanardi
764411
answered Nov 22 '18 at 21:06
cgiovanardicgiovanardi
764411
answered Nov 22 '18 at 21:06
cgiovanardicgiovanardi
764411
answered Nov 22 '18 at 21:06
cgiovanardicgiovanardi
764411
764411
add a comment |
add a comment |
$begingroup$
I'm assuming that you flipped right and left (since the real part of $300$ is clearly $300$). In general, to put a complex fraction in standard $x+yi$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:
$$ frac{a+bi}{c+di} ~~=~~ frac{a+bi}{c+di} cdot frac{c-di}{c-di} ~~=~~ frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$
Therefore, the real part of this fraction is
$$ frac{ac+bd}{c^2+d^2}. $$
In your case:
- $a=R277-X115$
- $b=R115+X277$
- $c=R+277$
- $d=X+115$
Plugging these into the formula above, we get
$$frac{[R^2277-RX115+R277^2-X(277cdot115)]+[RX115+X^2277+R115^2+X(277cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$
Cancelling terms in the numerator gives
$$=frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$
This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $115^2=13225$, and so on.
In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.
The resulting system is best solved by machine: here's the computation in WolframAlpha.
answered Dec 23 '18 at 4:57
aleph_twoaleph_two
27412
$endgroup$
add a comment |
$begingroup$
I'm assuming that you flipped right and left (since the real part of $300$ is clearly $300$). In general, to put a complex fraction in standard $x+yi$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:
$$ frac{a+bi}{c+di} ~~=~~ frac{a+bi}{c+di} cdot frac{c-di}{c-di} ~~=~~ frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$
Therefore, the real part of this fraction is
$$ frac{ac+bd}{c^2+d^2}. $$
In your case:
- $a=R277-X115$
- $b=R115+X277$
- $c=R+277$
- $d=X+115$
Plugging these into the formula above, we get
$$frac{[R^2277-RX115+R277^2-X(277cdot115)]+[RX115+X^2277+R115^2+X(277cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$
Cancelling terms in the numerator gives
$$=frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$
This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $115^2=13225$, and so on.
In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.
The resulting system is best solved by machine: here's the computation in WolframAlpha.
answered Dec 23 '18 at 4:57
aleph_twoaleph_two
27412
$endgroup$
add a comment |
$begingroup$
I'm assuming that you flipped right and left (since the real part of $300$ is clearly $300$). In general, to put a complex fraction in standard $x+yi$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:
$$ frac{a+bi}{c+di} ~~=~~ frac{a+bi}{c+di} cdot frac{c-di}{c-di} ~~=~~ frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$
Therefore, the real part of this fraction is
$$ frac{ac+bd}{c^2+d^2}. $$
In your case:
- $a=R277-X115$
- $b=R115+X277$
- $c=R+277$
- $d=X+115$
Plugging these into the formula above, we get
$$frac{[R^2277-RX115+R277^2-X(277cdot115)]+[RX115+X^2277+R115^2+X(277cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$
Cancelling terms in the numerator gives
$$=frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$
This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $115^2=13225$, and so on.
In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.
The resulting system is best solved by machine: here's the computation in WolframAlpha.
answered Dec 23 '18 at 4:57
aleph_twoaleph_two
27412
$endgroup$
I'm assuming that you flipped right and left (since the real part of $300$ is clearly $300$). In general, to put a complex fraction in standard $x+yi$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:
$$ frac{a+bi}{c+di} ~~=~~ frac{a+bi}{c+di} cdot frac{c-di}{c-di} ~~=~~ frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$
Therefore, the real part of this fraction is
$$ frac{ac+bd}{c^2+d^2}. $$
In your case:
- $a=R277-X115$
- $b=R115+X277$
- $c=R+277$
- $d=X+115$
Plugging these into the formula above, we get
$$frac{[R^2277-RX115+R277^2-X(277cdot115)]+[RX115+X^2277+R115^2+X(277cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$
Cancelling terms in the numerator gives
$$=frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$
This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $115^2=13225$, and so on.
In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.
The resulting system is best solved by machine: here's the computation in WolframAlpha.
answered Dec 23 '18 at 4:57
aleph_twoaleph_two
27412
answered Dec 23 '18 at 4:57
aleph_twoaleph_two
27412
answered Dec 23 '18 at 4:57
aleph_twoaleph_two
27412
answered Dec 23 '18 at 4:57
aleph_twoaleph_two
27412
27412
add a comment |
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$begingroup$
Well, $Re(300)=300$
$endgroup$
– gammatester
Nov 13 '18 at 10:39
1
$begingroup$
Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
$endgroup$
– saulspatz
Nov 13 '18 at 10:43