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 $












0












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I already figured out the answer to the first part zw but the rest is confusing me






calculus







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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
















0












$begingroup$


I already figured out the answer to the first part zw but the rest is confusing me






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


I already figured out the answer to the first part zw but the rest is confusing me






calculus







share|cite|improve this question









$endgroup$




I already figured out the answer to the first part zw but the rest is confusing me






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    youtube.com/watch?v=wlcDS7SKmj8 . There you have it, step by step.
    $endgroup$
    – Gil Astudillo
    Dec 10 '18 at 10:52





















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    youtube.com/watch?v=wlcDS7SKmj8 . There you have it, step by step.
    $endgroup$
    – Gil Astudillo
    Dec 10 '18 at 10:52


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    youtube.com/watch?v=wlcDS7SKmj8 . There you have it, step by step.
    $endgroup$
    – Gil Astudillo
    Dec 10 '18 at 10:52
















1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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













1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 22:15

























answered Dec 9 '18 at 21:42









Shubham JohriShubham Johri

5,092717




5,092717












  • $begingroup$
    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












  • $begingroup$
    $z=r_1e^{itheta_1}implies 1/z=frac1{r_1e^{itheta_1}}=frac1{r_1}cdot e^{i(-theta_1)}=r''e^{ibeta}=r''(cosbeta+isinbeta)$ (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


















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