Mixing
How to add real number and complex number in polar form
$begingroup$
I need to calculate
$$left| frac{3}{sqrt{20}} + i!cdot!frac{1}{sqrt{20}}!cdot!e^{i!cdot!frac{pi}{3}} right|$$
Is there a way to do it without turning the polar form into cartesian, multiply by $i$ and take magnitude of the resulting cartesian complex number? I've tried changing all the numbers to polar complex form but with different argument I'm not sure how to add.
|| is absolute value sign by the way.
complex-numbers
edited Jan 16 at 2:24
Larry
2,41331129
asked Jan 16 at 2:12
MinYoung KimMinYoung Kim
907
$endgroup$
add a comment |
$begingroup$
I need to calculate
$$left| frac{3}{sqrt{20}} + i!cdot!frac{1}{sqrt{20}}!cdot!e^{i!cdot!frac{pi}{3}} right|$$
Is there a way to do it without turning the polar form into cartesian, multiply by $i$ and take magnitude of the resulting cartesian complex number? I've tried changing all the numbers to polar complex form but with different argument I'm not sure how to add.
|| is absolute value sign by the way.
complex-numbers
edited Jan 16 at 2:24
Larry
2,41331129
asked Jan 16 at 2:12
MinYoung KimMinYoung Kim
907
$endgroup$
$begingroup$
Just change the polar part to cartesian and go about your business as usual. The geometric interpretation of the polar form may help you identify it as well.
$endgroup$
– CyclotomicField
Jan 16 at 2:29
$begingroup$
@CyclotomicField The question specifically asked if it were possible to do it without turning the polar into Cartesian form. So the answer is 'no', unless the arguments were the same or separated by $pi$.
$endgroup$
– Deepak
Jan 16 at 2:31
$begingroup$
@CyclotomicField Deepak's interpretation is correct, it feels like "brute force" to have to convert everything back to cartesian and distribute, was wondering if there's a more simpler way.
$endgroup$
– MinYoung Kim
Jan 16 at 2:38
$begingroup$
You don't need to. They are asking for the absolute value. So you just need to do that. Multiply by the complement and take the square root.
$endgroup$
– fleablood
Jan 16 at 3:07
add a comment |
$begingroup$
I need to calculate
$$left| frac{3}{sqrt{20}} + i!cdot!frac{1}{sqrt{20}}!cdot!e^{i!cdot!frac{pi}{3}} right|$$
Is there a way to do it without turning the polar form into cartesian, multiply by $i$ and take magnitude of the resulting cartesian complex number? I've tried changing all the numbers to polar complex form but with different argument I'm not sure how to add.
|| is absolute value sign by the way.
complex-numbers
edited Jan 16 at 2:24
Larry
2,41331129
asked Jan 16 at 2:12
MinYoung KimMinYoung Kim
907
$endgroup$
I need to calculate
$$left| frac{3}{sqrt{20}} + i!cdot!frac{1}{sqrt{20}}!cdot!e^{i!cdot!frac{pi}{3}} right|$$
Is there a way to do it without turning the polar form into cartesian, multiply by $i$ and take magnitude of the resulting cartesian complex number? I've tried changing all the numbers to polar complex form but with different argument I'm not sure how to add.
|| is absolute value sign by the way.
complex-numbers
complex-numbers
edited Jan 16 at 2:24
Larry
2,41331129
asked Jan 16 at 2:12
MinYoung KimMinYoung Kim
907
edited Jan 16 at 2:24
Larry
2,41331129
asked Jan 16 at 2:12
MinYoung KimMinYoung Kim
907
edited Jan 16 at 2:24
Larry
2,41331129
edited Jan 16 at 2:24
Larry
2,41331129
edited Jan 16 at 2:24
Larry
2,41331129
2,41331129
asked Jan 16 at 2:12
MinYoung KimMinYoung Kim
907
asked Jan 16 at 2:12
MinYoung KimMinYoung Kim
907
asked Jan 16 at 2:12
MinYoung KimMinYoung Kim
907
907
$begingroup$
Just change the polar part to cartesian and go about your business as usual. The geometric interpretation of the polar form may help you identify it as well.
$endgroup$
– CyclotomicField
Jan 16 at 2:29
$begingroup$
@CyclotomicField The question specifically asked if it were possible to do it without turning the polar into Cartesian form. So the answer is 'no', unless the arguments were the same or separated by $pi$.
$endgroup$
– Deepak
Jan 16 at 2:31
$begingroup$
@CyclotomicField Deepak's interpretation is correct, it feels like "brute force" to have to convert everything back to cartesian and distribute, was wondering if there's a more simpler way.
$endgroup$
– MinYoung Kim
Jan 16 at 2:38
$begingroup$
You don't need to. They are asking for the absolute value. So you just need to do that. Multiply by the complement and take the square root.
$endgroup$
– fleablood
Jan 16 at 3:07
add a comment |
$begingroup$
Just change the polar part to cartesian and go about your business as usual. The geometric interpretation of the polar form may help you identify it as well.
$endgroup$
– CyclotomicField
Jan 16 at 2:29
$begingroup$
@CyclotomicField The question specifically asked if it were possible to do it without turning the polar into Cartesian form. So the answer is 'no', unless the arguments were the same or separated by $pi$.
$endgroup$
– Deepak
Jan 16 at 2:31
$begingroup$
@CyclotomicField Deepak's interpretation is correct, it feels like "brute force" to have to convert everything back to cartesian and distribute, was wondering if there's a more simpler way.
$endgroup$
– MinYoung Kim
Jan 16 at 2:38
$begingroup$
You don't need to. They are asking for the absolute value. So you just need to do that. Multiply by the complement and take the square root.
$endgroup$
– fleablood
Jan 16 at 3:07
$begingroup$
Just change the polar part to cartesian and go about your business as usual. The geometric interpretation of the polar form may help you identify it as well.
$endgroup$
– CyclotomicField
Jan 16 at 2:29
$begingroup$
Just change the polar part to cartesian and go about your business as usual. The geometric interpretation of the polar form may help you identify it as well.
$endgroup$
– CyclotomicField
Jan 16 at 2:29
$begingroup$
@CyclotomicField The question specifically asked if it were possible to do it without turning the polar into Cartesian form. So the answer is 'no', unless the arguments were the same or separated by $pi$.
$endgroup$
– Deepak
Jan 16 at 2:31
$begingroup$
@CyclotomicField The question specifically asked if it were possible to do it without turning the polar into Cartesian form. So the answer is 'no', unless the arguments were the same or separated by $pi$.
$endgroup$
– Deepak
Jan 16 at 2:31
$begingroup$
@CyclotomicField Deepak's interpretation is correct, it feels like "brute force" to have to convert everything back to cartesian and distribute, was wondering if there's a more simpler way.
$endgroup$
– MinYoung Kim
Jan 16 at 2:38
$begingroup$
@CyclotomicField Deepak's interpretation is correct, it feels like "brute force" to have to convert everything back to cartesian and distribute, was wondering if there's a more simpler way.
$endgroup$
– MinYoung Kim
Jan 16 at 2:38
$begingroup$
You don't need to. They are asking for the absolute value. So you just need to do that. Multiply by the complement and take the square root.
$endgroup$
– fleablood
Jan 16 at 3:07
$begingroup$
You don't need to. They are asking for the absolute value. So you just need to do that. Multiply by the complement and take the square root.
$endgroup$
– fleablood
Jan 16 at 3:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's asking you not for the complex number but its absolute value.
The complement of $x + z$ where $x$ is real is $x + overline z$.
And the complement of $re^{itheta}$ is $re^{-itheta}$.
And $i*e^{itheta} = e^{ifrac pi 2}e^{itheta} = e^{i (theta +frac pi 2)}$.
So $|frac 3{sqrt {20}} + ifrac 1{sqrt{20}}e^{ifrac pi 3}| =$
$sqrt{(frac 3{sqrt {20}} +frac 1{sqrt{20}}e^{i(frac pi 3+frac pi 2)})(frac 3{sqrt {20}} +frac 1{sqrt{20}}e^{-i(frac pi 3+frac pi 2)})}=$
$sqrt{frac 9{20} + frac 3{20}(e^{ifrac {5pi}6} + e^{-ifrac {5pi}6}) + frac 1{20}}=$
$sqrt{frac {10}{20} + frac 3{20}*2cos frac {5pi}6}=$
$sqrt{frac 12 - frac {3sqrt 3}{20}}$
And I probably made an arithmetic error somewhere....
answered Jan 16 at 5:00
fleabloodfleablood
71.2k22686
$endgroup$
$begingroup$
Nope, that is the correct answer I got by turning everything into cartesian. The question actually asked for square of the magnitude, so without that square root, it makes things even simpler, $frac{1}{2} - frac{3sqrt{3}}{20}$ = .24, which is what I got. Thanks, didn't think about just multiplying its conjugate.
$endgroup$
– MinYoung Kim
Jan 16 at 16:34
$begingroup$
I say the $frac {sqrt{20}$ and the $3$ and the $1$s becoming nice I was expecting and nice clever negative or pythagorean triple simplifying out.
$endgroup$
– fleablood
Jan 16 at 16:50
add a comment |
$begingroup$
In general, it's only trivial to add (or subtract) two complex numbers in polar form if their arguments are the same or separated by $pi$.
When the arguments are the same, $ r_1e^{itheta} pm r_2{itheta} = (r_1 pm r_2)e^{itheta}$
When the arguments are separated by $pi$, $ r_1e^{itheta} pm r_2e^{i(theta pm pi)} = r_1e^{itheta} mp r_2e^{itheta} = (r_1 mp r_2)e^{itheta}$
answered Jan 16 at 2:34
DeepakDeepak
17k11536
$endgroup$
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
$begingroup$
It's asking you not for the complex number but its absolute value.
The complement of $x + z$ where $x$ is real is $x + overline z$.
And the complement of $re^{itheta}$ is $re^{-itheta}$.
And $i*e^{itheta} = e^{ifrac pi 2}e^{itheta} = e^{i (theta +frac pi 2)}$.
So $|frac 3{sqrt {20}} + ifrac 1{sqrt{20}}e^{ifrac pi 3}| =$
$sqrt{(frac 3{sqrt {20}} +frac 1{sqrt{20}}e^{i(frac pi 3+frac pi 2)})(frac 3{sqrt {20}} +frac 1{sqrt{20}}e^{-i(frac pi 3+frac pi 2)})}=$
$sqrt{frac 9{20} + frac 3{20}(e^{ifrac {5pi}6} + e^{-ifrac {5pi}6}) + frac 1{20}}=$
$sqrt{frac {10}{20} + frac 3{20}*2cos frac {5pi}6}=$
$sqrt{frac 12 - frac {3sqrt 3}{20}}$
And I probably made an arithmetic error somewhere....
answered Jan 16 at 5:00
fleabloodfleablood
71.2k22686
$endgroup$
$begingroup$
Nope, that is the correct answer I got by turning everything into cartesian. The question actually asked for square of the magnitude, so without that square root, it makes things even simpler, $frac{1}{2} - frac{3sqrt{3}}{20}$ = .24, which is what I got. Thanks, didn't think about just multiplying its conjugate.
$endgroup$
– MinYoung Kim
Jan 16 at 16:34
$begingroup$
I say the $frac {sqrt{20}$ and the $3$ and the $1$s becoming nice I was expecting and nice clever negative or pythagorean triple simplifying out.
$endgroup$
– fleablood
Jan 16 at 16:50
add a comment |
$begingroup$
It's asking you not for the complex number but its absolute value.
The complement of $x + z$ where $x$ is real is $x + overline z$.
And the complement of $re^{itheta}$ is $re^{-itheta}$.
And $i*e^{itheta} = e^{ifrac pi 2}e^{itheta} = e^{i (theta +frac pi 2)}$.
So $|frac 3{sqrt {20}} + ifrac 1{sqrt{20}}e^{ifrac pi 3}| =$
$sqrt{(frac 3{sqrt {20}} +frac 1{sqrt{20}}e^{i(frac pi 3+frac pi 2)})(frac 3{sqrt {20}} +frac 1{sqrt{20}}e^{-i(frac pi 3+frac pi 2)})}=$
$sqrt{frac 9{20} + frac 3{20}(e^{ifrac {5pi}6} + e^{-ifrac {5pi}6}) + frac 1{20}}=$
$sqrt{frac {10}{20} + frac 3{20}*2cos frac {5pi}6}=$
$sqrt{frac 12 - frac {3sqrt 3}{20}}$
And I probably made an arithmetic error somewhere....
answered Jan 16 at 5:00
fleabloodfleablood
71.2k22686
$endgroup$
$begingroup$
Nope, that is the correct answer I got by turning everything into cartesian. The question actually asked for square of the magnitude, so without that square root, it makes things even simpler, $frac{1}{2} - frac{3sqrt{3}}{20}$ = .24, which is what I got. Thanks, didn't think about just multiplying its conjugate.
$endgroup$
– MinYoung Kim
Jan 16 at 16:34
$begingroup$
I say the $frac {sqrt{20}$ and the $3$ and the $1$s becoming nice I was expecting and nice clever negative or pythagorean triple simplifying out.
$endgroup$
– fleablood
Jan 16 at 16:50
add a comment |
$begingroup$
It's asking you not for the complex number but its absolute value.
The complement of $x + z$ where $x$ is real is $x + overline z$.
And the complement of $re^{itheta}$ is $re^{-itheta}$.
And $i*e^{itheta} = e^{ifrac pi 2}e^{itheta} = e^{i (theta +frac pi 2)}$.
So $|frac 3{sqrt {20}} + ifrac 1{sqrt{20}}e^{ifrac pi 3}| =$
$sqrt{(frac 3{sqrt {20}} +frac 1{sqrt{20}}e^{i(frac pi 3+frac pi 2)})(frac 3{sqrt {20}} +frac 1{sqrt{20}}e^{-i(frac pi 3+frac pi 2)})}=$
$sqrt{frac 9{20} + frac 3{20}(e^{ifrac {5pi}6} + e^{-ifrac {5pi}6}) + frac 1{20}}=$
$sqrt{frac {10}{20} + frac 3{20}*2cos frac {5pi}6}=$
$sqrt{frac 12 - frac {3sqrt 3}{20}}$
And I probably made an arithmetic error somewhere....
answered Jan 16 at 5:00
fleabloodfleablood
71.2k22686
$endgroup$
It's asking you not for the complex number but its absolute value.
The complement of $x + z$ where $x$ is real is $x + overline z$.
And the complement of $re^{itheta}$ is $re^{-itheta}$.
And $i*e^{itheta} = e^{ifrac pi 2}e^{itheta} = e^{i (theta +frac pi 2)}$.
So $|frac 3{sqrt {20}} + ifrac 1{sqrt{20}}e^{ifrac pi 3}| =$
$sqrt{(frac 3{sqrt {20}} +frac 1{sqrt{20}}e^{i(frac pi 3+frac pi 2)})(frac 3{sqrt {20}} +frac 1{sqrt{20}}e^{-i(frac pi 3+frac pi 2)})}=$
$sqrt{frac 9{20} + frac 3{20}(e^{ifrac {5pi}6} + e^{-ifrac {5pi}6}) + frac 1{20}}=$
$sqrt{frac {10}{20} + frac 3{20}*2cos frac {5pi}6}=$
$sqrt{frac 12 - frac {3sqrt 3}{20}}$
And I probably made an arithmetic error somewhere....
answered Jan 16 at 5:00
fleabloodfleablood
71.2k22686
answered Jan 16 at 5:00
fleabloodfleablood
71.2k22686
answered Jan 16 at 5:00
fleabloodfleablood
71.2k22686
answered Jan 16 at 5:00
fleabloodfleablood
71.2k22686
71.2k22686
$begingroup$
Nope, that is the correct answer I got by turning everything into cartesian. The question actually asked for square of the magnitude, so without that square root, it makes things even simpler, $frac{1}{2} - frac{3sqrt{3}}{20}$ = .24, which is what I got. Thanks, didn't think about just multiplying its conjugate.
$endgroup$
– MinYoung Kim
Jan 16 at 16:34
$begingroup$
I say the $frac {sqrt{20}$ and the $3$ and the $1$s becoming nice I was expecting and nice clever negative or pythagorean triple simplifying out.
$endgroup$
– fleablood
Jan 16 at 16:50
add a comment |
$begingroup$
Nope, that is the correct answer I got by turning everything into cartesian. The question actually asked for square of the magnitude, so without that square root, it makes things even simpler, $frac{1}{2} - frac{3sqrt{3}}{20}$ = .24, which is what I got. Thanks, didn't think about just multiplying its conjugate.
$endgroup$
– MinYoung Kim
Jan 16 at 16:34
$begingroup$
I say the $frac {sqrt{20}$ and the $3$ and the $1$s becoming nice I was expecting and nice clever negative or pythagorean triple simplifying out.
$endgroup$
– fleablood
Jan 16 at 16:50
$begingroup$
Nope, that is the correct answer I got by turning everything into cartesian. The question actually asked for square of the magnitude, so without that square root, it makes things even simpler, $frac{1}{2} - frac{3sqrt{3}}{20}$ = .24, which is what I got. Thanks, didn't think about just multiplying its conjugate.
$endgroup$
– MinYoung Kim
Jan 16 at 16:34
$begingroup$
Nope, that is the correct answer I got by turning everything into cartesian. The question actually asked for square of the magnitude, so without that square root, it makes things even simpler, $frac{1}{2} - frac{3sqrt{3}}{20}$ = .24, which is what I got. Thanks, didn't think about just multiplying its conjugate.
$endgroup$
– MinYoung Kim
Jan 16 at 16:34
$begingroup$
I say the $frac {sqrt{20}$ and the $3$ and the $1$s becoming nice I was expecting and nice clever negative or pythagorean triple simplifying out.
$endgroup$
– fleablood
Jan 16 at 16:50
$begingroup$
I say the $frac {sqrt{20}$ and the $3$ and the $1$s becoming nice I was expecting and nice clever negative or pythagorean triple simplifying out.
$endgroup$
– fleablood
Jan 16 at 16:50
add a comment |
$begingroup$
In general, it's only trivial to add (or subtract) two complex numbers in polar form if their arguments are the same or separated by $pi$.
When the arguments are the same, $ r_1e^{itheta} pm r_2{itheta} = (r_1 pm r_2)e^{itheta}$
When the arguments are separated by $pi$, $ r_1e^{itheta} pm r_2e^{i(theta pm pi)} = r_1e^{itheta} mp r_2e^{itheta} = (r_1 mp r_2)e^{itheta}$
answered Jan 16 at 2:34
DeepakDeepak
17k11536
$endgroup$
add a comment |
$begingroup$
In general, it's only trivial to add (or subtract) two complex numbers in polar form if their arguments are the same or separated by $pi$.
When the arguments are the same, $ r_1e^{itheta} pm r_2{itheta} = (r_1 pm r_2)e^{itheta}$
When the arguments are separated by $pi$, $ r_1e^{itheta} pm r_2e^{i(theta pm pi)} = r_1e^{itheta} mp r_2e^{itheta} = (r_1 mp r_2)e^{itheta}$
answered Jan 16 at 2:34
DeepakDeepak
17k11536
$endgroup$
add a comment |
$begingroup$
In general, it's only trivial to add (or subtract) two complex numbers in polar form if their arguments are the same or separated by $pi$.
When the arguments are the same, $ r_1e^{itheta} pm r_2{itheta} = (r_1 pm r_2)e^{itheta}$
When the arguments are separated by $pi$, $ r_1e^{itheta} pm r_2e^{i(theta pm pi)} = r_1e^{itheta} mp r_2e^{itheta} = (r_1 mp r_2)e^{itheta}$
answered Jan 16 at 2:34
DeepakDeepak
17k11536
$endgroup$
In general, it's only trivial to add (or subtract) two complex numbers in polar form if their arguments are the same or separated by $pi$.
When the arguments are the same, $ r_1e^{itheta} pm r_2{itheta} = (r_1 pm r_2)e^{itheta}$
When the arguments are separated by $pi$, $ r_1e^{itheta} pm r_2e^{i(theta pm pi)} = r_1e^{itheta} mp r_2e^{itheta} = (r_1 mp r_2)e^{itheta}$
answered Jan 16 at 2:34
DeepakDeepak
17k11536
answered Jan 16 at 2:34
DeepakDeepak
17k11536
answered Jan 16 at 2:34
DeepakDeepak
17k11536
answered Jan 16 at 2:34
DeepakDeepak
17k11536
17k11536
add a comment |
add a comment |
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$begingroup$
Just change the polar part to cartesian and go about your business as usual. The geometric interpretation of the polar form may help you identify it as well.
$endgroup$
– CyclotomicField
Jan 16 at 2:29
$begingroup$
@CyclotomicField The question specifically asked if it were possible to do it without turning the polar into Cartesian form. So the answer is 'no', unless the arguments were the same or separated by $pi$.
$endgroup$
– Deepak
Jan 16 at 2:31
$begingroup$
@CyclotomicField Deepak's interpretation is correct, it feels like "brute force" to have to convert everything back to cartesian and distribute, was wondering if there's a more simpler way.
$endgroup$
– MinYoung Kim
Jan 16 at 2:38
$begingroup$
You don't need to. They are asking for the absolute value. So you just need to do that. Multiply by the complement and take the square root.
$endgroup$
– fleablood
Jan 16 at 3:07