Is $|zbar{z}^{-1}|=1,zinmathbb{C}$? [closed]
-1
$bar{z}$ is the conjugation of $z$.
Thank you so much
complex-numbers
asked Nov 29 at 14:41
RM777
1928
closed as off-topic by Michael Burr, Jyrki Lahtonen, Davide Giraudo, Leucippus, KReiser Nov 30 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Michael Burr, Jyrki Lahtonen, Davide Giraudo, Leucippus, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
-1
$bar{z}$ is the conjugation of $z$.
Thank you so much
complex-numbers
asked Nov 29 at 14:41
RM777
1928
closed as off-topic by Michael Burr, Jyrki Lahtonen, Davide Giraudo, Leucippus, KReiser Nov 30 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Michael Burr, Jyrki Lahtonen, Davide Giraudo, Leucippus, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
1
What have you tried?
– Michael Burr
Nov 29 at 14:43
1
The result I got was if $z = a+bi$ the Statement would be equivalent to $(a+bi)(a-bi)^{-1}$ which is equivalent to (according to my computation) $frac{a^2-b^2}{a^2+b^2}+2frac{ab}{a^2+b^2}i$
– RM777
Nov 29 at 14:47
1
You are almost there, compute the modulus of that number !
– nicomezi
Nov 29 at 14:49
2
It's easier not to bother with real and imaginary parts, if you know that $|zw|=|z||w|$ and $|overline{z}| = |z|$. Of course you need $z ne 0$ for your left side to be defined.
– Robert Israel
Nov 29 at 14:51
My next question Building upon this result is if $zin S_1$ (Unit circle) is $zbar{z}^{-1}=z^2$?
– RM777
Nov 29 at 15:09
add a comment |
-1
-1
-1
$bar{z}$ is the conjugation of $z$.
Thank you so much
complex-numbers
asked Nov 29 at 14:41
RM777
1928
$bar{z}$ is the conjugation of $z$.
Thank you so much
complex-numbers
complex-numbers
asked Nov 29 at 14:41
RM777
1928
asked Nov 29 at 14:41
RM777
1928
asked Nov 29 at 14:41
RM777
1928
asked Nov 29 at 14:41
RM777
1928
asked Nov 29 at 14:41
RM777
1928
1928
closed as off-topic by Michael Burr, Jyrki Lahtonen, Davide Giraudo, Leucippus, KReiser Nov 30 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Michael Burr, Jyrki Lahtonen, Davide Giraudo, Leucippus, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Michael Burr, Jyrki Lahtonen, Davide Giraudo, Leucippus, KReiser Nov 30 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Michael Burr, Jyrki Lahtonen, Davide Giraudo, Leucippus, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
1
What have you tried?
– Michael Burr
Nov 29 at 14:43
1
The result I got was if $z = a+bi$ the Statement would be equivalent to $(a+bi)(a-bi)^{-1}$ which is equivalent to (according to my computation) $frac{a^2-b^2}{a^2+b^2}+2frac{ab}{a^2+b^2}i$
– RM777
Nov 29 at 14:47
1
You are almost there, compute the modulus of that number !
– nicomezi
Nov 29 at 14:49
2
It's easier not to bother with real and imaginary parts, if you know that $|zw|=|z||w|$ and $|overline{z}| = |z|$. Of course you need $z ne 0$ for your left side to be defined.
– Robert Israel
Nov 29 at 14:51
My next question Building upon this result is if $zin S_1$ (Unit circle) is $zbar{z}^{-1}=z^2$?
– RM777
Nov 29 at 15:09
add a comment |
1
What have you tried?
– Michael Burr
Nov 29 at 14:43
1
The result I got was if $z = a+bi$ the Statement would be equivalent to $(a+bi)(a-bi)^{-1}$ which is equivalent to (according to my computation) $frac{a^2-b^2}{a^2+b^2}+2frac{ab}{a^2+b^2}i$
– RM777
Nov 29 at 14:47
1
You are almost there, compute the modulus of that number !
– nicomezi
Nov 29 at 14:49
2
It's easier not to bother with real and imaginary parts, if you know that $|zw|=|z||w|$ and $|overline{z}| = |z|$. Of course you need $z ne 0$ for your left side to be defined.
– Robert Israel
Nov 29 at 14:51
My next question Building upon this result is if $zin S_1$ (Unit circle) is $zbar{z}^{-1}=z^2$?
– RM777
Nov 29 at 15:09
1
1
What have you tried?
– Michael Burr
Nov 29 at 14:43
What have you tried?
– Michael Burr
Nov 29 at 14:43
1
1
The result I got was if $z = a+bi$ the Statement would be equivalent to $(a+bi)(a-bi)^{-1}$ which is equivalent to (according to my computation) $frac{a^2-b^2}{a^2+b^2}+2frac{ab}{a^2+b^2}i$
– RM777
Nov 29 at 14:47
The result I got was if $z = a+bi$ the Statement would be equivalent to $(a+bi)(a-bi)^{-1}$ which is equivalent to (according to my computation) $frac{a^2-b^2}{a^2+b^2}+2frac{ab}{a^2+b^2}i$
– RM777
Nov 29 at 14:47
1
1
You are almost there, compute the modulus of that number !
– nicomezi
Nov 29 at 14:49
You are almost there, compute the modulus of that number !
– nicomezi
Nov 29 at 14:49
2
2
It's easier not to bother with real and imaginary parts, if you know that $|zw|=|z||w|$ and $|overline{z}| = |z|$. Of course you need $z ne 0$ for your left side to be defined.
– Robert Israel
Nov 29 at 14:51
It's easier not to bother with real and imaginary parts, if you know that $|zw|=|z||w|$ and $|overline{z}| = |z|$. Of course you need $z ne 0$ for your left side to be defined.
– Robert Israel
Nov 29 at 14:51
My next question Building upon this result is if $zin S_1$ (Unit circle) is $zbar{z}^{-1}=z^2$?
– RM777
Nov 29 at 15:09
My next question Building upon this result is if $zin S_1$ (Unit circle) is $zbar{z}^{-1}=z^2$?
– RM777
Nov 29 at 15:09
add a comment |
2 Answers
2
active
oldest
votes
0
HINT
Use that with $bar z ne 0$
$$|zbar{z}^{-1}|=|z||bar{z}^{-1}|=frac{|z|}{|bar{z}|}$$
answered Nov 29 at 14:51
gimusi
1
add a comment |
0
Note that $|z|=|bar{z}| space forall z in mathbb{C}$ and $left| frac{z_1}{z_2} right| = frac{|z_1|}{|z_2|} space forall z_1, z_2 in mathbb{C}, z_2 ne 0$
answered Nov 29 at 14:52
gandalf61
7,663623
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
HINT
Use that with $bar z ne 0$
$$|zbar{z}^{-1}|=|z||bar{z}^{-1}|=frac{|z|}{|bar{z}|}$$
answered Nov 29 at 14:51
gimusi
1
add a comment |
0
HINT
Use that with $bar z ne 0$
$$|zbar{z}^{-1}|=|z||bar{z}^{-1}|=frac{|z|}{|bar{z}|}$$
answered Nov 29 at 14:51
gimusi
1
add a comment |
0
0
0
HINT
Use that with $bar z ne 0$
$$|zbar{z}^{-1}|=|z||bar{z}^{-1}|=frac{|z|}{|bar{z}|}$$
answered Nov 29 at 14:51
gimusi
1
HINT
Use that with $bar z ne 0$
$$|zbar{z}^{-1}|=|z||bar{z}^{-1}|=frac{|z|}{|bar{z}|}$$
answered Nov 29 at 14:51
gimusi
1
answered Nov 29 at 14:51
gimusi
1
answered Nov 29 at 14:51
gimusi
1
answered Nov 29 at 14:51
gimusi
1
1
add a comment |
add a comment |
0
Note that $|z|=|bar{z}| space forall z in mathbb{C}$ and $left| frac{z_1}{z_2} right| = frac{|z_1|}{|z_2|} space forall z_1, z_2 in mathbb{C}, z_2 ne 0$
answered Nov 29 at 14:52
gandalf61
7,663623
add a comment |
0
Note that $|z|=|bar{z}| space forall z in mathbb{C}$ and $left| frac{z_1}{z_2} right| = frac{|z_1|}{|z_2|} space forall z_1, z_2 in mathbb{C}, z_2 ne 0$
answered Nov 29 at 14:52
gandalf61
7,663623
add a comment |
0
0
0
Note that $|z|=|bar{z}| space forall z in mathbb{C}$ and $left| frac{z_1}{z_2} right| = frac{|z_1|}{|z_2|} space forall z_1, z_2 in mathbb{C}, z_2 ne 0$
answered Nov 29 at 14:52
gandalf61
7,663623
Note that $|z|=|bar{z}| space forall z in mathbb{C}$ and $left| frac{z_1}{z_2} right| = frac{|z_1|}{|z_2|} space forall z_1, z_2 in mathbb{C}, z_2 ne 0$
answered Nov 29 at 14:52
gandalf61
7,663623
answered Nov 29 at 14:52
gandalf61
7,663623
answered Nov 29 at 14:52
gandalf61
7,663623
answered Nov 29 at 14:52
gandalf61
7,663623
7,663623
add a comment |
add a comment |
1
What have you tried?
– Michael Burr
Nov 29 at 14:43
1
The result I got was if $z = a+bi$ the Statement would be equivalent to $(a+bi)(a-bi)^{-1}$ which is equivalent to (according to my computation) $frac{a^2-b^2}{a^2+b^2}+2frac{ab}{a^2+b^2}i$
– RM777
Nov 29 at 14:47
1
You are almost there, compute the modulus of that number !
– nicomezi
Nov 29 at 14:49
2
It's easier not to bother with real and imaginary parts, if you know that $|zw|=|z||w|$ and $|overline{z}| = |z|$. Of course you need $z ne 0$ for your left side to be defined.
– Robert Israel
Nov 29 at 14:51
My next question Building upon this result is if $zin S_1$ (Unit circle) is $zbar{z}^{-1}=z^2$?
– RM777
Nov 29 at 15:09