Find polar forms for zw, z/w, and 1/z by first putting z and w into polar form. $ z=2sqrt{3}-2i, w=-1+i $
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I already figured out the answer to the first part zw but the rest is confusing me
calculus
asked Dec 9 '18 at 21:33
mysticemmamysticemma
1
$endgroup$
add a comment |
$begingroup$
I already figured out the answer to the first part zw but the rest is confusing me
calculus
asked Dec 9 '18 at 21:33
mysticemmamysticemma
1
$endgroup$
$begingroup$
Welcome to Maths SX! And what did you obtain?
$endgroup$
– Bernard
Dec 9 '18 at 21:36
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How did you "figure out the answer to the first part?" The other parts should not be a challenge.
$endgroup$
– Mark Viola
Dec 9 '18 at 21:36
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well I ended up getting 4sqrt2(cos(7pi/4)+isin(7pi/4)) by multiplying the two roots together and then adding the two angles in radians but I can't seem to calculate the answer for the other two because I'm not sure how those rules must translate. I just needed some help with that
$endgroup$
– mysticemma
Dec 9 '18 at 21:44
$begingroup$
I've also been stuck on this for like 2hrs now but I must be missing something major I would like to know how to do this easily and efficiently
$endgroup$
– mysticemma
Dec 9 '18 at 21:46
add a comment |
$begingroup$
I already figured out the answer to the first part zw but the rest is confusing me
calculus
asked Dec 9 '18 at 21:33
mysticemmamysticemma
1
$endgroup$
I already figured out the answer to the first part zw but the rest is confusing me
calculus
calculus
asked Dec 9 '18 at 21:33
mysticemmamysticemma
1
asked Dec 9 '18 at 21:33
mysticemmamysticemma
1
asked Dec 9 '18 at 21:33
mysticemmamysticemma
1
asked Dec 9 '18 at 21:33
mysticemmamysticemma
1
asked Dec 9 '18 at 21:33
mysticemmamysticemma
1
1
$begingroup$
Welcome to Maths SX! And what did you obtain?
$endgroup$
– Bernard
Dec 9 '18 at 21:36
$begingroup$
How did you "figure out the answer to the first part?" The other parts should not be a challenge.
$endgroup$
– Mark Viola
Dec 9 '18 at 21:36
$begingroup$
well I ended up getting 4sqrt2(cos(7pi/4)+isin(7pi/4)) by multiplying the two roots together and then adding the two angles in radians but I can't seem to calculate the answer for the other two because I'm not sure how those rules must translate. I just needed some help with that
$endgroup$
– mysticemma
Dec 9 '18 at 21:44
$begingroup$
I've also been stuck on this for like 2hrs now but I must be missing something major I would like to know how to do this easily and efficiently
$endgroup$
– mysticemma
Dec 9 '18 at 21:46
add a comment |
$begingroup$
Welcome to Maths SX! And what did you obtain?
$endgroup$
– Bernard
Dec 9 '18 at 21:36
$begingroup$
How did you "figure out the answer to the first part?" The other parts should not be a challenge.
$endgroup$
– Mark Viola
Dec 9 '18 at 21:36
$begingroup$
well I ended up getting 4sqrt2(cos(7pi/4)+isin(7pi/4)) by multiplying the two roots together and then adding the two angles in radians but I can't seem to calculate the answer for the other two because I'm not sure how those rules must translate. I just needed some help with that
$endgroup$
– mysticemma
Dec 9 '18 at 21:44
$begingroup$
I've also been stuck on this for like 2hrs now but I must be missing something major I would like to know how to do this easily and efficiently
$endgroup$
– mysticemma
Dec 9 '18 at 21:46
$begingroup$
Welcome to Maths SX! And what did you obtain?
$endgroup$
– Bernard
Dec 9 '18 at 21:36
$begingroup$
Welcome to Maths SX! And what did you obtain?
$endgroup$
– Bernard
Dec 9 '18 at 21:36
$begingroup$
How did you "figure out the answer to the first part?" The other parts should not be a challenge.
$endgroup$
– Mark Viola
Dec 9 '18 at 21:36
$begingroup$
How did you "figure out the answer to the first part?" The other parts should not be a challenge.
$endgroup$
– Mark Viola
Dec 9 '18 at 21:36
$begingroup$
well I ended up getting 4sqrt2(cos(7pi/4)+isin(7pi/4)) by multiplying the two roots together and then adding the two angles in radians but I can't seem to calculate the answer for the other two because I'm not sure how those rules must translate. I just needed some help with that
$endgroup$
– mysticemma
Dec 9 '18 at 21:44
$begingroup$
well I ended up getting 4sqrt2(cos(7pi/4)+isin(7pi/4)) by multiplying the two roots together and then adding the two angles in radians but I can't seem to calculate the answer for the other two because I'm not sure how those rules must translate. I just needed some help with that
$endgroup$
– mysticemma
Dec 9 '18 at 21:44
$begingroup$
I've also been stuck on this for like 2hrs now but I must be missing something major I would like to know how to do this easily and efficiently
$endgroup$
– mysticemma
Dec 9 '18 at 21:46
$begingroup$
I've also been stuck on this for like 2hrs now but I must be missing something major I would like to know how to do this easily and efficiently
$endgroup$
– mysticemma
Dec 9 '18 at 21:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
https://en.wikipedia.org/wiki/Complex_number#Polar_form . Look at this page.
You need to calculate the norm and the angument to get the polar form of $z$ and $w$, and then do the operations that you want.
answered Dec 9 '18 at 21:47
Gil AstudilloGil Astudillo
315
$endgroup$
$begingroup$
youtube.com/watch?v=wlcDS7SKmj8 . There you have it, step by step.
$endgroup$
– Gil Astudillo
Dec 10 '18 at 10:52
add a comment |
$begingroup$
Hint
$z=r_1e^{itheta_1}, w=r_2e^{itheta_2}\z/w=(frac{r_1}{r_2})cdot e^{i({theta_1-theta_2})}=r'e^{ialpha}\1/z=(frac1{r_1}).e^{i(-theta_1)}=r''e^{ibeta}$
You need to take care that $alpha,beta$ lie in the domain of the principal argument.
Edit
One way to write the polar form of a complex number $z$ is $|z|(cos Arg(z)+isin Arg(z))$, while another way to write this is $|z|e^{iArg(z)}$, where $|z|$ denotes the amplitude of $z$ and $Arg(z)$ its principal argument. This is because $e^{iArg(z)}=cos Arg(z)+isin Arg(z)$, which comes from the Euler's formula.
answered Dec 9 '18 at 21:42
Shubham JohriShubham Johri
5,192717
$endgroup$
$begingroup$
How do you obtain the reciprocal of z in trigonometric form? I have the second part I believe in trigonometric form now, as 2sqrt2(cos(13pi/12)+isin(13pi/6)).
$endgroup$
– mysticemma
Dec 9 '18 at 22:00
$begingroup$
$z=r_1e^{itheta_1}implies 1/z=frac1{r_1e^{itheta_1}}=frac1{r_1}cdot e^{i(-theta_1)}=r''e^{ibeta}=r''(cosbeta+isinbeta)$ (Euler's formula) where $r''=frac1{r_1}, beta=-theta_1$. Check whether $beta$ is in the domain of principal argument.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:04
$begingroup$
I have edited the answer to include a little introduction to the Euler's Formula. Please refer to the link.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:17
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
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oldest
votes
$begingroup$
https://en.wikipedia.org/wiki/Complex_number#Polar_form . Look at this page.
You need to calculate the norm and the angument to get the polar form of $z$ and $w$, and then do the operations that you want.
answered Dec 9 '18 at 21:47
Gil AstudilloGil Astudillo
315
$endgroup$
$begingroup$
youtube.com/watch?v=wlcDS7SKmj8 . There you have it, step by step.
$endgroup$
– Gil Astudillo
Dec 10 '18 at 10:52
add a comment |
$begingroup$
https://en.wikipedia.org/wiki/Complex_number#Polar_form . Look at this page.
You need to calculate the norm and the angument to get the polar form of $z$ and $w$, and then do the operations that you want.
answered Dec 9 '18 at 21:47
Gil AstudilloGil Astudillo
315
$endgroup$
$begingroup$
youtube.com/watch?v=wlcDS7SKmj8 . There you have it, step by step.
$endgroup$
– Gil Astudillo
Dec 10 '18 at 10:52
add a comment |
$begingroup$
https://en.wikipedia.org/wiki/Complex_number#Polar_form . Look at this page.
You need to calculate the norm and the angument to get the polar form of $z$ and $w$, and then do the operations that you want.
answered Dec 9 '18 at 21:47
Gil AstudilloGil Astudillo
315
$endgroup$
https://en.wikipedia.org/wiki/Complex_number#Polar_form . Look at this page.
You need to calculate the norm and the angument to get the polar form of $z$ and $w$, and then do the operations that you want.
answered Dec 9 '18 at 21:47
Gil AstudilloGil Astudillo
315
answered Dec 9 '18 at 21:47
Gil AstudilloGil Astudillo
315
answered Dec 9 '18 at 21:47
Gil AstudilloGil Astudillo
315
answered Dec 9 '18 at 21:47
Gil AstudilloGil Astudillo
315
315
$begingroup$
youtube.com/watch?v=wlcDS7SKmj8 . There you have it, step by step.
$endgroup$
– Gil Astudillo
Dec 10 '18 at 10:52
add a comment |
$begingroup$
youtube.com/watch?v=wlcDS7SKmj8 . There you have it, step by step.
$endgroup$
– Gil Astudillo
Dec 10 '18 at 10:52
$begingroup$
youtube.com/watch?v=wlcDS7SKmj8 . There you have it, step by step.
$endgroup$
– Gil Astudillo
Dec 10 '18 at 10:52
$begingroup$
youtube.com/watch?v=wlcDS7SKmj8 . There you have it, step by step.
$endgroup$
– Gil Astudillo
Dec 10 '18 at 10:52
add a comment |
$begingroup$
Hint
$z=r_1e^{itheta_1}, w=r_2e^{itheta_2}\z/w=(frac{r_1}{r_2})cdot e^{i({theta_1-theta_2})}=r'e^{ialpha}\1/z=(frac1{r_1}).e^{i(-theta_1)}=r''e^{ibeta}$
You need to take care that $alpha,beta$ lie in the domain of the principal argument.
Edit
One way to write the polar form of a complex number $z$ is $|z|(cos Arg(z)+isin Arg(z))$, while another way to write this is $|z|e^{iArg(z)}$, where $|z|$ denotes the amplitude of $z$ and $Arg(z)$ its principal argument. This is because $e^{iArg(z)}=cos Arg(z)+isin Arg(z)$, which comes from the Euler's formula.
answered Dec 9 '18 at 21:42
Shubham JohriShubham Johri
5,192717
$endgroup$
$begingroup$
How do you obtain the reciprocal of z in trigonometric form? I have the second part I believe in trigonometric form now, as 2sqrt2(cos(13pi/12)+isin(13pi/6)).
$endgroup$
– mysticemma
Dec 9 '18 at 22:00
$begingroup$
$z=r_1e^{itheta_1}implies 1/z=frac1{r_1e^{itheta_1}}=frac1{r_1}cdot e^{i(-theta_1)}=r''e^{ibeta}=r''(cosbeta+isinbeta)$ (Euler's formula) where $r''=frac1{r_1}, beta=-theta_1$. Check whether $beta$ is in the domain of principal argument.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:04
$begingroup$
I have edited the answer to include a little introduction to the Euler's Formula. Please refer to the link.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:17
add a comment |
$begingroup$
Hint
$z=r_1e^{itheta_1}, w=r_2e^{itheta_2}\z/w=(frac{r_1}{r_2})cdot e^{i({theta_1-theta_2})}=r'e^{ialpha}\1/z=(frac1{r_1}).e^{i(-theta_1)}=r''e^{ibeta}$
You need to take care that $alpha,beta$ lie in the domain of the principal argument.
Edit
One way to write the polar form of a complex number $z$ is $|z|(cos Arg(z)+isin Arg(z))$, while another way to write this is $|z|e^{iArg(z)}$, where $|z|$ denotes the amplitude of $z$ and $Arg(z)$ its principal argument. This is because $e^{iArg(z)}=cos Arg(z)+isin Arg(z)$, which comes from the Euler's formula.
answered Dec 9 '18 at 21:42
Shubham JohriShubham Johri
5,192717
$endgroup$
$begingroup$
How do you obtain the reciprocal of z in trigonometric form? I have the second part I believe in trigonometric form now, as 2sqrt2(cos(13pi/12)+isin(13pi/6)).
$endgroup$
– mysticemma
Dec 9 '18 at 22:00
$begingroup$
$z=r_1e^{itheta_1}implies 1/z=frac1{r_1e^{itheta_1}}=frac1{r_1}cdot e^{i(-theta_1)}=r''e^{ibeta}=r''(cosbeta+isinbeta)$ (Euler's formula) where $r''=frac1{r_1}, beta=-theta_1$. Check whether $beta$ is in the domain of principal argument.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:04
$begingroup$
I have edited the answer to include a little introduction to the Euler's Formula. Please refer to the link.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:17
add a comment |
$begingroup$
Hint
$z=r_1e^{itheta_1}, w=r_2e^{itheta_2}\z/w=(frac{r_1}{r_2})cdot e^{i({theta_1-theta_2})}=r'e^{ialpha}\1/z=(frac1{r_1}).e^{i(-theta_1)}=r''e^{ibeta}$
You need to take care that $alpha,beta$ lie in the domain of the principal argument.
Edit
One way to write the polar form of a complex number $z$ is $|z|(cos Arg(z)+isin Arg(z))$, while another way to write this is $|z|e^{iArg(z)}$, where $|z|$ denotes the amplitude of $z$ and $Arg(z)$ its principal argument. This is because $e^{iArg(z)}=cos Arg(z)+isin Arg(z)$, which comes from the Euler's formula.
answered Dec 9 '18 at 21:42
Shubham JohriShubham Johri
5,192717
$endgroup$
Hint
$z=r_1e^{itheta_1}, w=r_2e^{itheta_2}\z/w=(frac{r_1}{r_2})cdot e^{i({theta_1-theta_2})}=r'e^{ialpha}\1/z=(frac1{r_1}).e^{i(-theta_1)}=r''e^{ibeta}$
You need to take care that $alpha,beta$ lie in the domain of the principal argument.
Edit
One way to write the polar form of a complex number $z$ is $|z|(cos Arg(z)+isin Arg(z))$, while another way to write this is $|z|e^{iArg(z)}$, where $|z|$ denotes the amplitude of $z$ and $Arg(z)$ its principal argument. This is because $e^{iArg(z)}=cos Arg(z)+isin Arg(z)$, which comes from the Euler's formula.
answered Dec 9 '18 at 21:42
Shubham JohriShubham Johri
5,192717
edited Dec 9 '18 at 22:15
answered Dec 9 '18 at 21:42
Shubham JohriShubham Johri
5,192717
answered Dec 9 '18 at 21:42
Shubham JohriShubham Johri
5,192717
answered Dec 9 '18 at 21:42
Shubham JohriShubham Johri
5,192717
5,192717
$begingroup$
How do you obtain the reciprocal of z in trigonometric form? I have the second part I believe in trigonometric form now, as 2sqrt2(cos(13pi/12)+isin(13pi/6)).
$endgroup$
– mysticemma
Dec 9 '18 at 22:00
$begingroup$
$z=r_1e^{itheta_1}implies 1/z=frac1{r_1e^{itheta_1}}=frac1{r_1}cdot e^{i(-theta_1)}=r''e^{ibeta}=r''(cosbeta+isinbeta)$ (Euler's formula) where $r''=frac1{r_1}, beta=-theta_1$. Check whether $beta$ is in the domain of principal argument.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:04
$begingroup$
I have edited the answer to include a little introduction to the Euler's Formula. Please refer to the link.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:17
add a comment |
$begingroup$
How do you obtain the reciprocal of z in trigonometric form? I have the second part I believe in trigonometric form now, as 2sqrt2(cos(13pi/12)+isin(13pi/6)).
$endgroup$
– mysticemma
Dec 9 '18 at 22:00
$begingroup$
$z=r_1e^{itheta_1}implies 1/z=frac1{r_1e^{itheta_1}}=frac1{r_1}cdot e^{i(-theta_1)}=r''e^{ibeta}=r''(cosbeta+isinbeta)$ (Euler's formula) where $r''=frac1{r_1}, beta=-theta_1$. Check whether $beta$ is in the domain of principal argument.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:04
$begingroup$
I have edited the answer to include a little introduction to the Euler's Formula. Please refer to the link.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:17
$begingroup$
How do you obtain the reciprocal of z in trigonometric form? I have the second part I believe in trigonometric form now, as 2sqrt2(cos(13pi/12)+isin(13pi/6)).
$endgroup$
– mysticemma
Dec 9 '18 at 22:00
$begingroup$
How do you obtain the reciprocal of z in trigonometric form? I have the second part I believe in trigonometric form now, as 2sqrt2(cos(13pi/12)+isin(13pi/6)).
$endgroup$
– mysticemma
Dec 9 '18 at 22:00
$begingroup$
$z=r_1e^{itheta_1}implies 1/z=frac1{r_1e^{itheta_1}}=frac1{r_1}cdot e^{i(-theta_1)}=r''e^{ibeta}=r''(cosbeta+isinbeta)$ (Euler's formula) where $r''=frac1{r_1}, beta=-theta_1$. Check whether $beta$ is in the domain of principal argument.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:04
$begingroup$
$z=r_1e^{itheta_1}implies 1/z=frac1{r_1e^{itheta_1}}=frac1{r_1}cdot e^{i(-theta_1)}=r''e^{ibeta}=r''(cosbeta+isinbeta)$ (Euler's formula) where $r''=frac1{r_1}, beta=-theta_1$. Check whether $beta$ is in the domain of principal argument.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:04
$begingroup$
I have edited the answer to include a little introduction to the Euler's Formula. Please refer to the link.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:17
$begingroup$
I have edited the answer to include a little introduction to the Euler's Formula. Please refer to the link.
$endgroup$
– Shubham Johri
Dec 9 '18 at 22:17
add a comment |
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$begingroup$
Welcome to Maths SX! And what did you obtain?
$endgroup$
– Bernard
Dec 9 '18 at 21:36
$begingroup$
How did you "figure out the answer to the first part?" The other parts should not be a challenge.
$endgroup$
– Mark Viola
Dec 9 '18 at 21:36
$begingroup$
well I ended up getting 4sqrt2(cos(7pi/4)+isin(7pi/4)) by multiplying the two roots together and then adding the two angles in radians but I can't seem to calculate the answer for the other two because I'm not sure how those rules must translate. I just needed some help with that
$endgroup$
– mysticemma
Dec 9 '18 at 21:44
$begingroup$
I've also been stuck on this for like 2hrs now but I must be missing something major I would like to know how to do this easily and efficiently
$endgroup$
– mysticemma
Dec 9 '18 at 21:46