Mixing
Multiplying complex numbers in polar form?
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Could someone explain why you multiply the lengths and add the angles when multiplying polar coordinates?
I tried multiplying the polar forms ($r_1left(costheta_1 + isintheta_1right)cdot r_2left(costheta_2 + isintheta_2right)$), and expanding/factoring the result, and end up multiplying the lengths but can't seem to come to an equation where you add the angles.
complex-analysis complex-numbers
edited May 14 '12 at 0:45
Michael Hardy
1
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add a comment |
$begingroup$
Could someone explain why you multiply the lengths and add the angles when multiplying polar coordinates?
I tried multiplying the polar forms ($r_1left(costheta_1 + isintheta_1right)cdot r_2left(costheta_2 + isintheta_2right)$), and expanding/factoring the result, and end up multiplying the lengths but can't seem to come to an equation where you add the angles.
complex-analysis complex-numbers
edited May 14 '12 at 0:45
Michael Hardy
1
$endgroup$
1
$begingroup$
Are you familiar with the identities for $cos(theta_1+theta_2)$ and $sin(theta_1+theta_2)$?
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– Gerry Myerson
May 14 '12 at 0:45
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@GerryMyerson No I was not. I plugged them in, and arrived at the correct equation. Thanks!
$endgroup$
– user26649
May 14 '12 at 0:50
add a comment |
$begingroup$
Could someone explain why you multiply the lengths and add the angles when multiplying polar coordinates?
I tried multiplying the polar forms ($r_1left(costheta_1 + isintheta_1right)cdot r_2left(costheta_2 + isintheta_2right)$), and expanding/factoring the result, and end up multiplying the lengths but can't seem to come to an equation where you add the angles.
complex-analysis complex-numbers
edited May 14 '12 at 0:45
Michael Hardy
1
$endgroup$
Could someone explain why you multiply the lengths and add the angles when multiplying polar coordinates?
I tried multiplying the polar forms ($r_1left(costheta_1 + isintheta_1right)cdot r_2left(costheta_2 + isintheta_2right)$), and expanding/factoring the result, and end up multiplying the lengths but can't seem to come to an equation where you add the angles.
complex-analysis complex-numbers
complex-analysis complex-numbers
edited May 14 '12 at 0:45
Michael Hardy
1
edited May 14 '12 at 0:45
Michael Hardy
1
edited May 14 '12 at 0:45
Michael Hardy
1
edited May 14 '12 at 0:45
Michael Hardy
1
edited May 14 '12 at 0:45
Michael Hardy
1
1
asked May 14 '12 at 0:40
user26649
1
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Are you familiar with the identities for $cos(theta_1+theta_2)$ and $sin(theta_1+theta_2)$?
$endgroup$
– Gerry Myerson
May 14 '12 at 0:45
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@GerryMyerson No I was not. I plugged them in, and arrived at the correct equation. Thanks!
$endgroup$
– user26649
May 14 '12 at 0:50
add a comment |
1
$begingroup$
Are you familiar with the identities for $cos(theta_1+theta_2)$ and $sin(theta_1+theta_2)$?
$endgroup$
– Gerry Myerson
May 14 '12 at 0:45
$begingroup$
@GerryMyerson No I was not. I plugged them in, and arrived at the correct equation. Thanks!
$endgroup$
– user26649
May 14 '12 at 0:50
1
1
$begingroup$
Are you familiar with the identities for $cos(theta_1+theta_2)$ and $sin(theta_1+theta_2)$?
$endgroup$
– Gerry Myerson
May 14 '12 at 0:45
$begingroup$
Are you familiar with the identities for $cos(theta_1+theta_2)$ and $sin(theta_1+theta_2)$?
$endgroup$
– Gerry Myerson
May 14 '12 at 0:45
$begingroup$
@GerryMyerson No I was not. I plugged them in, and arrived at the correct equation. Thanks!
$endgroup$
– user26649
May 14 '12 at 0:50
$begingroup$
@GerryMyerson No I was not. I plugged them in, and arrived at the correct equation. Thanks!
$endgroup$
– user26649
May 14 '12 at 0:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By multiplying things out as usual, you get
$$[r_1(costheta_1 + isintheta_1)][r_2(costheta_2 + isintheta_2)] = r_1r_2(costheta_1costheta_2 - sintheta_1sintheta_2 + i[sintheta_1costheta_2 + sintheta_2costheta_1]).$$
Now you want to use the trig identities $costheta_1costheta_2 - sintheta_1sintheta_2 = cos(theta_1 + theta_2)$ and $sintheta_1costheta_2 + sintheta_2costheta_1 = sin(theta_1 + theta_2)$ to conclude that this is in fact $$r_1r_2[cos(theta_1 + theta_2) + isin(theta_1 + theta_2)].$$
answered May 14 '12 at 0:45
froggiefroggie
8,55211741
$endgroup$
3
$begingroup$
Well since that's exactly what I was halfway through typing, I certainly deserves my vote!
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– The Chaz 2.0
May 14 '12 at 0:47
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My thoughts, precisely.
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– Cameron Buie
May 14 '12 at 0:57
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@Froggie Thanks!
$endgroup$
– user26649
May 14 '12 at 1:05
add a comment |
$begingroup$
It might be useful to write the numbers as
$$z_1= nu e^{itheta}$$
$$z_2= mu e^{ipsi}$$
Then one has
$$z_1cdot z_2 =nu mu cdot e^{(psi+theta)i}$$
This representation stems from Euler's formula
$$e^{i theta}=cos theta+isin theta$$
which I suspect you haven't been told about yet.
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
By multiplying things out as usual, you get
$$[r_1(costheta_1 + isintheta_1)][r_2(costheta_2 + isintheta_2)] = r_1r_2(costheta_1costheta_2 - sintheta_1sintheta_2 + i[sintheta_1costheta_2 + sintheta_2costheta_1]).$$
Now you want to use the trig identities $costheta_1costheta_2 - sintheta_1sintheta_2 = cos(theta_1 + theta_2)$ and $sintheta_1costheta_2 + sintheta_2costheta_1 = sin(theta_1 + theta_2)$ to conclude that this is in fact $$r_1r_2[cos(theta_1 + theta_2) + isin(theta_1 + theta_2)].$$
answered May 14 '12 at 0:45
froggiefroggie
8,55211741
$endgroup$
3
$begingroup$
Well since that's exactly what I was halfway through typing, I certainly deserves my vote!
$endgroup$
– The Chaz 2.0
May 14 '12 at 0:47
$begingroup$
My thoughts, precisely.
$endgroup$
– Cameron Buie
May 14 '12 at 0:57
$begingroup$
@Froggie Thanks!
$endgroup$
– user26649
May 14 '12 at 1:05
add a comment |
$begingroup$
By multiplying things out as usual, you get
$$[r_1(costheta_1 + isintheta_1)][r_2(costheta_2 + isintheta_2)] = r_1r_2(costheta_1costheta_2 - sintheta_1sintheta_2 + i[sintheta_1costheta_2 + sintheta_2costheta_1]).$$
Now you want to use the trig identities $costheta_1costheta_2 - sintheta_1sintheta_2 = cos(theta_1 + theta_2)$ and $sintheta_1costheta_2 + sintheta_2costheta_1 = sin(theta_1 + theta_2)$ to conclude that this is in fact $$r_1r_2[cos(theta_1 + theta_2) + isin(theta_1 + theta_2)].$$
answered May 14 '12 at 0:45
froggiefroggie
8,55211741
$endgroup$
3
$begingroup$
Well since that's exactly what I was halfway through typing, I certainly deserves my vote!
$endgroup$
– The Chaz 2.0
May 14 '12 at 0:47
$begingroup$
My thoughts, precisely.
$endgroup$
– Cameron Buie
May 14 '12 at 0:57
$begingroup$
@Froggie Thanks!
$endgroup$
– user26649
May 14 '12 at 1:05
add a comment |
$begingroup$
By multiplying things out as usual, you get
$$[r_1(costheta_1 + isintheta_1)][r_2(costheta_2 + isintheta_2)] = r_1r_2(costheta_1costheta_2 - sintheta_1sintheta_2 + i[sintheta_1costheta_2 + sintheta_2costheta_1]).$$
Now you want to use the trig identities $costheta_1costheta_2 - sintheta_1sintheta_2 = cos(theta_1 + theta_2)$ and $sintheta_1costheta_2 + sintheta_2costheta_1 = sin(theta_1 + theta_2)$ to conclude that this is in fact $$r_1r_2[cos(theta_1 + theta_2) + isin(theta_1 + theta_2)].$$
answered May 14 '12 at 0:45
froggiefroggie
8,55211741
$endgroup$
By multiplying things out as usual, you get
$$[r_1(costheta_1 + isintheta_1)][r_2(costheta_2 + isintheta_2)] = r_1r_2(costheta_1costheta_2 - sintheta_1sintheta_2 + i[sintheta_1costheta_2 + sintheta_2costheta_1]).$$
Now you want to use the trig identities $costheta_1costheta_2 - sintheta_1sintheta_2 = cos(theta_1 + theta_2)$ and $sintheta_1costheta_2 + sintheta_2costheta_1 = sin(theta_1 + theta_2)$ to conclude that this is in fact $$r_1r_2[cos(theta_1 + theta_2) + isin(theta_1 + theta_2)].$$
answered May 14 '12 at 0:45
froggiefroggie
8,55211741
answered May 14 '12 at 0:45
froggiefroggie
8,55211741
answered May 14 '12 at 0:45
froggiefroggie
8,55211741
answered May 14 '12 at 0:45
froggiefroggie
8,55211741
8,55211741
3
$begingroup$
Well since that's exactly what I was halfway through typing, I certainly deserves my vote!
$endgroup$
– The Chaz 2.0
May 14 '12 at 0:47
$begingroup$
My thoughts, precisely.
$endgroup$
– Cameron Buie
May 14 '12 at 0:57
$begingroup$
@Froggie Thanks!
$endgroup$
– user26649
May 14 '12 at 1:05
add a comment |
3
$begingroup$
Well since that's exactly what I was halfway through typing, I certainly deserves my vote!
$endgroup$
– The Chaz 2.0
May 14 '12 at 0:47
$begingroup$
My thoughts, precisely.
$endgroup$
– Cameron Buie
May 14 '12 at 0:57
$begingroup$
@Froggie Thanks!
$endgroup$
– user26649
May 14 '12 at 1:05
3
3
$begingroup$
Well since that's exactly what I was halfway through typing, I certainly deserves my vote!
$endgroup$
– The Chaz 2.0
May 14 '12 at 0:47
$begingroup$
Well since that's exactly what I was halfway through typing, I certainly deserves my vote!
$endgroup$
– The Chaz 2.0
May 14 '12 at 0:47
$begingroup$
My thoughts, precisely.
$endgroup$
– Cameron Buie
May 14 '12 at 0:57
$begingroup$
My thoughts, precisely.
$endgroup$
– Cameron Buie
May 14 '12 at 0:57
$begingroup$
@Froggie Thanks!
$endgroup$
– user26649
May 14 '12 at 1:05
$begingroup$
@Froggie Thanks!
$endgroup$
– user26649
May 14 '12 at 1:05
add a comment |
$begingroup$
It might be useful to write the numbers as
$$z_1= nu e^{itheta}$$
$$z_2= mu e^{ipsi}$$
Then one has
$$z_1cdot z_2 =nu mu cdot e^{(psi+theta)i}$$
This representation stems from Euler's formula
$$e^{i theta}=cos theta+isin theta$$
which I suspect you haven't been told about yet.
$endgroup$
add a comment |
$begingroup$
It might be useful to write the numbers as
$$z_1= nu e^{itheta}$$
$$z_2= mu e^{ipsi}$$
Then one has
$$z_1cdot z_2 =nu mu cdot e^{(psi+theta)i}$$
This representation stems from Euler's formula
$$e^{i theta}=cos theta+isin theta$$
which I suspect you haven't been told about yet.
$endgroup$
add a comment |
$begingroup$
It might be useful to write the numbers as
$$z_1= nu e^{itheta}$$
$$z_2= mu e^{ipsi}$$
Then one has
$$z_1cdot z_2 =nu mu cdot e^{(psi+theta)i}$$
This representation stems from Euler's formula
$$e^{i theta}=cos theta+isin theta$$
which I suspect you haven't been told about yet.
$endgroup$
It might be useful to write the numbers as
$$z_1= nu e^{itheta}$$
$$z_2= mu e^{ipsi}$$
Then one has
$$z_1cdot z_2 =nu mu cdot e^{(psi+theta)i}$$
This representation stems from Euler's formula
$$e^{i theta}=cos theta+isin theta$$
which I suspect you haven't been told about yet.
answered May 14 '12 at 1:10
community wiki
Pedro Tamaroff
add a comment |
add a comment |
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1
$begingroup$
Are you familiar with the identities for $cos(theta_1+theta_2)$ and $sin(theta_1+theta_2)$?
$endgroup$
– Gerry Myerson
May 14 '12 at 0:45
$begingroup$
@GerryMyerson No I was not. I plugged them in, and arrived at the correct equation. Thanks!
$endgroup$
– user26649
May 14 '12 at 0:50