Prove $frac{bar z}{z^2}-frac{z}{bar z^2}$ is a pure imaginary number











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I have come across the intriguing statement that $frac{bar z}{z^2}-frac{z}{bar z^2}$ is an imaginary number, but somehow - no matter what I try - I cannot seem to get to that result (I have tried a long list of operations, but all of the outcomes still had the 'a' of 'a+ib' or 'a-ib' in it in the end). Can anybody shed some light on this?




Thank you!






algebra-precalculus complex-numbers







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  • 3




    You mean pure imaginary? Well...take it's complex conjugate.
    – lulu
    Nov 23 at 15:43















up vote
0
down vote

favorite













I have come across the intriguing statement that $frac{bar z}{z^2}-frac{z}{bar z^2}$ is an imaginary number, but somehow - no matter what I try - I cannot seem to get to that result (I have tried a long list of operations, but all of the outcomes still had the 'a' of 'a+ib' or 'a-ib' in it in the end). Can anybody shed some light on this?




Thank you!






algebra-precalculus complex-numbers







share|cite|improve this question




















  • 3




    You mean pure imaginary? Well...take it's complex conjugate.
    – lulu
    Nov 23 at 15:43













up vote
0
down vote

favorite









up vote
0
down vote

favorite












I have come across the intriguing statement that $frac{bar z}{z^2}-frac{z}{bar z^2}$ is an imaginary number, but somehow - no matter what I try - I cannot seem to get to that result (I have tried a long list of operations, but all of the outcomes still had the 'a' of 'a+ib' or 'a-ib' in it in the end). Can anybody shed some light on this?




Thank you!






algebra-precalculus complex-numbers







share|cite|improve this question
















I have come across the intriguing statement that $frac{bar z}{z^2}-frac{z}{bar z^2}$ is an imaginary number, but somehow - no matter what I try - I cannot seem to get to that result (I have tried a long list of operations, but all of the outcomes still had the 'a' of 'a+ib' or 'a-ib' in it in the end). Can anybody shed some light on this?




Thank you!






algebra-precalculus complex-numbers




algebra-precalculus complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 16:43









Math Lover

13.6k31435




13.6k31435










asked Nov 23 at 15:40









Pregunto

122




122








  • 3




    You mean pure imaginary? Well...take it's complex conjugate.
    – lulu
    Nov 23 at 15:43














  • 3




    You mean pure imaginary? Well...take it's complex conjugate.
    – lulu
    Nov 23 at 15:43








3




3




You mean pure imaginary? Well...take it's complex conjugate.
– lulu
Nov 23 at 15:43




You mean pure imaginary? Well...take it's complex conjugate.
– lulu
Nov 23 at 15:43










2 Answers
2






active

oldest

votes

















up vote
2
down vote













Let
$$frac{overline{z}}{z^2} = a+ib.$$
Then
$$overline{left(frac{overline{z}}{z^2}right)} =frac{z}{overline{z}^2}= a-ib.$$
Consequently,
$$frac{overline{z}}{z^2} - frac{z}{overline{z}^2} = ?$$






share|cite|improve this answer























  • Thanks! Could you just explain the rule on which the second equation is based? How do you know that z(bar)/z^2 = z/z(bar)^2? I cannot find it anywhere in my book...
    – Pregunto
    Nov 23 at 16:17






  • 1




    I think the rule is $overline{xcdot y} = overline{x} cdot overline{y}$ and $overline{overline{x}}=x$. You can easily verify them.
    – Babelfish
    Nov 23 at 16:21












  • @Pregunto Babelfish has mentioned the rules.
    – Math Lover
    Nov 23 at 16:28


















up vote
1
down vote













Note that $frac{overline{z}}{z^2}-frac{z}{overline{z}^2} = frac{overline{z}^3-z^3}{z^2overline{z}^2}$ (1). You should know that $zoverline{z} = lvert zrvert in mathbb{R}$. By consequence, the denominator of (1) is real. To prove that your expression is imaginary now comes down to proving that the numerator of (1) is imaginary. From here on it should be a lot easier. The conclusion of the proof is in the spoiler.




It is now useful to write $z=a+ib$, with $a,binmathbb{R}$. The numerator of (1) then becomes $(a^3-ia^2b-ab^2+ib^3)-(a^3+ia^2b-ab^2-ib^3) = -2ia^2b+2ib^3$, which is completely imaginary. We are done.







share|cite|improve this answer





















  • So that's the thing: I got that far, but then I assumed that the real denominator keeps the number real. So that was a wrong assumption?
    – Pregunto
    Nov 23 at 16:30










  • Consider $frac{ia}{b}$. The denominator is real and the enumerator is imaginary. This number is of course equal to $ifrac{a}{b}$, which is imaginary ($i$ times the real number $frac{a}{b}$). That assumption was wrong. It might be useful to think about how you came to that assumption, it could clear things up and lead to a better understanding.
    – Aldoggen
    Nov 23 at 16:37










  • Now I get it… thanks!
    – Pregunto
    Nov 23 at 16:58











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

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active

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active

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up vote
2
down vote













Let
$$frac{overline{z}}{z^2} = a+ib.$$
Then
$$overline{left(frac{overline{z}}{z^2}right)} =frac{z}{overline{z}^2}= a-ib.$$
Consequently,
$$frac{overline{z}}{z^2} - frac{z}{overline{z}^2} = ?$$






share|cite|improve this answer























  • Thanks! Could you just explain the rule on which the second equation is based? How do you know that z(bar)/z^2 = z/z(bar)^2? I cannot find it anywhere in my book...
    – Pregunto
    Nov 23 at 16:17






  • 1




    I think the rule is $overline{xcdot y} = overline{x} cdot overline{y}$ and $overline{overline{x}}=x$. You can easily verify them.
    – Babelfish
    Nov 23 at 16:21












  • @Pregunto Babelfish has mentioned the rules.
    – Math Lover
    Nov 23 at 16:28















up vote
2
down vote













Let
$$frac{overline{z}}{z^2} = a+ib.$$
Then
$$overline{left(frac{overline{z}}{z^2}right)} =frac{z}{overline{z}^2}= a-ib.$$
Consequently,
$$frac{overline{z}}{z^2} - frac{z}{overline{z}^2} = ?$$






share|cite|improve this answer























  • Thanks! Could you just explain the rule on which the second equation is based? How do you know that z(bar)/z^2 = z/z(bar)^2? I cannot find it anywhere in my book...
    – Pregunto
    Nov 23 at 16:17






  • 1




    I think the rule is $overline{xcdot y} = overline{x} cdot overline{y}$ and $overline{overline{x}}=x$. You can easily verify them.
    – Babelfish
    Nov 23 at 16:21












  • @Pregunto Babelfish has mentioned the rules.
    – Math Lover
    Nov 23 at 16:28













up vote
2
down vote










up vote
2
down vote









Let
$$frac{overline{z}}{z^2} = a+ib.$$
Then
$$overline{left(frac{overline{z}}{z^2}right)} =frac{z}{overline{z}^2}= a-ib.$$
Consequently,
$$frac{overline{z}}{z^2} - frac{z}{overline{z}^2} = ?$$






share|cite|improve this answer














Let
$$frac{overline{z}}{z^2} = a+ib.$$
Then
$$overline{left(frac{overline{z}}{z^2}right)} =frac{z}{overline{z}^2}= a-ib.$$
Consequently,
$$frac{overline{z}}{z^2} - frac{z}{overline{z}^2} = ?$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 23 at 16:44









Rahul

32.9k466162




32.9k466162










answered Nov 23 at 15:58









Math Lover

13.6k31435




13.6k31435












  • Thanks! Could you just explain the rule on which the second equation is based? How do you know that z(bar)/z^2 = z/z(bar)^2? I cannot find it anywhere in my book...
    – Pregunto
    Nov 23 at 16:17






  • 1




    I think the rule is $overline{xcdot y} = overline{x} cdot overline{y}$ and $overline{overline{x}}=x$. You can easily verify them.
    – Babelfish
    Nov 23 at 16:21












  • @Pregunto Babelfish has mentioned the rules.
    – Math Lover
    Nov 23 at 16:28


















  • Thanks! Could you just explain the rule on which the second equation is based? How do you know that z(bar)/z^2 = z/z(bar)^2? I cannot find it anywhere in my book...
    – Pregunto
    Nov 23 at 16:17






  • 1




    I think the rule is $overline{xcdot y} = overline{x} cdot overline{y}$ and $overline{overline{x}}=x$. You can easily verify them.
    – Babelfish
    Nov 23 at 16:21












  • @Pregunto Babelfish has mentioned the rules.
    – Math Lover
    Nov 23 at 16:28
















Thanks! Could you just explain the rule on which the second equation is based? How do you know that z(bar)/z^2 = z/z(bar)^2? I cannot find it anywhere in my book...
– Pregunto
Nov 23 at 16:17




Thanks! Could you just explain the rule on which the second equation is based? How do you know that z(bar)/z^2 = z/z(bar)^2? I cannot find it anywhere in my book...
– Pregunto
Nov 23 at 16:17




1




1




I think the rule is $overline{xcdot y} = overline{x} cdot overline{y}$ and $overline{overline{x}}=x$. You can easily verify them.
– Babelfish
Nov 23 at 16:21






I think the rule is $overline{xcdot y} = overline{x} cdot overline{y}$ and $overline{overline{x}}=x$. You can easily verify them.
– Babelfish
Nov 23 at 16:21














@Pregunto Babelfish has mentioned the rules.
– Math Lover
Nov 23 at 16:28




@Pregunto Babelfish has mentioned the rules.
– Math Lover
Nov 23 at 16:28










up vote
1
down vote













Note that $frac{overline{z}}{z^2}-frac{z}{overline{z}^2} = frac{overline{z}^3-z^3}{z^2overline{z}^2}$ (1). You should know that $zoverline{z} = lvert zrvert in mathbb{R}$. By consequence, the denominator of (1) is real. To prove that your expression is imaginary now comes down to proving that the numerator of (1) is imaginary. From here on it should be a lot easier. The conclusion of the proof is in the spoiler.




It is now useful to write $z=a+ib$, with $a,binmathbb{R}$. The numerator of (1) then becomes $(a^3-ia^2b-ab^2+ib^3)-(a^3+ia^2b-ab^2-ib^3) = -2ia^2b+2ib^3$, which is completely imaginary. We are done.







share|cite|improve this answer





















  • So that's the thing: I got that far, but then I assumed that the real denominator keeps the number real. So that was a wrong assumption?
    – Pregunto
    Nov 23 at 16:30










  • Consider $frac{ia}{b}$. The denominator is real and the enumerator is imaginary. This number is of course equal to $ifrac{a}{b}$, which is imaginary ($i$ times the real number $frac{a}{b}$). That assumption was wrong. It might be useful to think about how you came to that assumption, it could clear things up and lead to a better understanding.
    – Aldoggen
    Nov 23 at 16:37










  • Now I get it… thanks!
    – Pregunto
    Nov 23 at 16:58















up vote
1
down vote













Note that $frac{overline{z}}{z^2}-frac{z}{overline{z}^2} = frac{overline{z}^3-z^3}{z^2overline{z}^2}$ (1). You should know that $zoverline{z} = lvert zrvert in mathbb{R}$. By consequence, the denominator of (1) is real. To prove that your expression is imaginary now comes down to proving that the numerator of (1) is imaginary. From here on it should be a lot easier. The conclusion of the proof is in the spoiler.




It is now useful to write $z=a+ib$, with $a,binmathbb{R}$. The numerator of (1) then becomes $(a^3-ia^2b-ab^2+ib^3)-(a^3+ia^2b-ab^2-ib^3) = -2ia^2b+2ib^3$, which is completely imaginary. We are done.







share|cite|improve this answer





















  • So that's the thing: I got that far, but then I assumed that the real denominator keeps the number real. So that was a wrong assumption?
    – Pregunto
    Nov 23 at 16:30










  • Consider $frac{ia}{b}$. The denominator is real and the enumerator is imaginary. This number is of course equal to $ifrac{a}{b}$, which is imaginary ($i$ times the real number $frac{a}{b}$). That assumption was wrong. It might be useful to think about how you came to that assumption, it could clear things up and lead to a better understanding.
    – Aldoggen
    Nov 23 at 16:37










  • Now I get it… thanks!
    – Pregunto
    Nov 23 at 16:58













up vote
1
down vote










up vote
1
down vote









Note that $frac{overline{z}}{z^2}-frac{z}{overline{z}^2} = frac{overline{z}^3-z^3}{z^2overline{z}^2}$ (1). You should know that $zoverline{z} = lvert zrvert in mathbb{R}$. By consequence, the denominator of (1) is real. To prove that your expression is imaginary now comes down to proving that the numerator of (1) is imaginary. From here on it should be a lot easier. The conclusion of the proof is in the spoiler.




It is now useful to write $z=a+ib$, with $a,binmathbb{R}$. The numerator of (1) then becomes $(a^3-ia^2b-ab^2+ib^3)-(a^3+ia^2b-ab^2-ib^3) = -2ia^2b+2ib^3$, which is completely imaginary. We are done.







share|cite|improve this answer












Note that $frac{overline{z}}{z^2}-frac{z}{overline{z}^2} = frac{overline{z}^3-z^3}{z^2overline{z}^2}$ (1). You should know that $zoverline{z} = lvert zrvert in mathbb{R}$. By consequence, the denominator of (1) is real. To prove that your expression is imaginary now comes down to proving that the numerator of (1) is imaginary. From here on it should be a lot easier. The conclusion of the proof is in the spoiler.




It is now useful to write $z=a+ib$, with $a,binmathbb{R}$. The numerator of (1) then becomes $(a^3-ia^2b-ab^2+ib^3)-(a^3+ia^2b-ab^2-ib^3) = -2ia^2b+2ib^3$, which is completely imaginary. We are done.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 23 at 16:25









Aldoggen

514




514












  • So that's the thing: I got that far, but then I assumed that the real denominator keeps the number real. So that was a wrong assumption?
    – Pregunto
    Nov 23 at 16:30










  • Consider $frac{ia}{b}$. The denominator is real and the enumerator is imaginary. This number is of course equal to $ifrac{a}{b}$, which is imaginary ($i$ times the real number $frac{a}{b}$). That assumption was wrong. It might be useful to think about how you came to that assumption, it could clear things up and lead to a better understanding.
    – Aldoggen
    Nov 23 at 16:37










  • Now I get it… thanks!
    – Pregunto
    Nov 23 at 16:58


















  • So that's the thing: I got that far, but then I assumed that the real denominator keeps the number real. So that was a wrong assumption?
    – Pregunto
    Nov 23 at 16:30










  • Consider $frac{ia}{b}$. The denominator is real and the enumerator is imaginary. This number is of course equal to $ifrac{a}{b}$, which is imaginary ($i$ times the real number $frac{a}{b}$). That assumption was wrong. It might be useful to think about how you came to that assumption, it could clear things up and lead to a better understanding.
    – Aldoggen
    Nov 23 at 16:37










  • Now I get it… thanks!
    – Pregunto
    Nov 23 at 16:58
















So that's the thing: I got that far, but then I assumed that the real denominator keeps the number real. So that was a wrong assumption?
– Pregunto
Nov 23 at 16:30




So that's the thing: I got that far, but then I assumed that the real denominator keeps the number real. So that was a wrong assumption?
– Pregunto
Nov 23 at 16:30












Consider $frac{ia}{b}$. The denominator is real and the enumerator is imaginary. This number is of course equal to $ifrac{a}{b}$, which is imaginary ($i$ times the real number $frac{a}{b}$). That assumption was wrong. It might be useful to think about how you came to that assumption, it could clear things up and lead to a better understanding.
– Aldoggen
Nov 23 at 16:37




Consider $frac{ia}{b}$. The denominator is real and the enumerator is imaginary. This number is of course equal to $ifrac{a}{b}$, which is imaginary ($i$ times the real number $frac{a}{b}$). That assumption was wrong. It might be useful to think about how you came to that assumption, it could clear things up and lead to a better understanding.
– Aldoggen
Nov 23 at 16:37












Now I get it… thanks!
– Pregunto
Nov 23 at 16:58




Now I get it… thanks!
– Pregunto
Nov 23 at 16:58


















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