Show that $|{|2overline{z}+5 |(sqrt2 - i)} | = sqrt3 |2z+5|$, where z is a complex number.
up vote
-1
down vote
favorite
Show that $|{|2overline{z}+5 |(sqrt2 - i)} | = sqrt3 |2z+5|$, where z is a complex number.
And $overline{z}$ is complex conjugate of $z$.
And $i$ is iota.
I'm proceeding by considering $z=x+iy$
But I just get stuck at different results approaching different ways.
Please help.
complex-analysis complex-numbers
asked Nov 21 at 13:40
Kaustuv Sawarn
465
add a comment |
up vote
-1
down vote
favorite
Show that $|{|2overline{z}+5 |(sqrt2 - i)} | = sqrt3 |2z+5|$, where z is a complex number.
And $overline{z}$ is complex conjugate of $z$.
And $i$ is iota.
I'm proceeding by considering $z=x+iy$
But I just get stuck at different results approaching different ways.
Please help.
complex-analysis complex-numbers
asked Nov 21 at 13:40
Kaustuv Sawarn
465
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
Nov 21 at 13:46
@PeterSzilas I didn't understand, how can you bring LHS as $sqrt3|2overline{z}+5|$ . I mean how did you proceed with the LHS to get that!?
– Kaustuv Sawarn
Nov 21 at 13:57
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 at 16:12
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 at 16:29
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 at 16:35
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Show that $|{|2overline{z}+5 |(sqrt2 - i)} | = sqrt3 |2z+5|$, where z is a complex number.
And $overline{z}$ is complex conjugate of $z$.
And $i$ is iota.
I'm proceeding by considering $z=x+iy$
But I just get stuck at different results approaching different ways.
Please help.
complex-analysis complex-numbers
asked Nov 21 at 13:40
Kaustuv Sawarn
465
Show that $|{|2overline{z}+5 |(sqrt2 - i)} | = sqrt3 |2z+5|$, where z is a complex number.
And $overline{z}$ is complex conjugate of $z$.
And $i$ is iota.
I'm proceeding by considering $z=x+iy$
But I just get stuck at different results approaching different ways.
Please help.
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Nov 21 at 13:40
Kaustuv Sawarn
465
asked Nov 21 at 13:40
Kaustuv Sawarn
465
asked Nov 21 at 13:40
Kaustuv Sawarn
465
asked Nov 21 at 13:40
Kaustuv Sawarn
465
asked Nov 21 at 13:40
Kaustuv Sawarn
465
465
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
Nov 21 at 13:46
@PeterSzilas I didn't understand, how can you bring LHS as $sqrt3|2overline{z}+5|$ . I mean how did you proceed with the LHS to get that!?
– Kaustuv Sawarn
Nov 21 at 13:57
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 at 16:12
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 at 16:29
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 at 16:35
add a comment |
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
Nov 21 at 13:46
@PeterSzilas I didn't understand, how can you bring LHS as $sqrt3|2overline{z}+5|$ . I mean how did you proceed with the LHS to get that!?
– Kaustuv Sawarn
Nov 21 at 13:57
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 at 16:12
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 at 16:29
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 at 16:35
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
Nov 21 at 13:46
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
Nov 21 at 13:46
@PeterSzilas I didn't understand, how can you bring LHS as $sqrt3|2overline{z}+5|$ . I mean how did you proceed with the LHS to get that!?
– Kaustuv Sawarn
Nov 21 at 13:57
@PeterSzilas I didn't understand, how can you bring LHS as $sqrt3|2overline{z}+5|$ . I mean how did you proceed with the LHS to get that!?
– Kaustuv Sawarn
Nov 21 at 13:57
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 at 16:12
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 at 16:12
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 at 16:29
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 at 16:29
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 at 16:35
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 at 16:35
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
LHS is $||2overline{z}+5|cdot(sqrt2 - i)|$; and $a=|2overline{z}+5| geq 0$. Then
$$||2overline{z}+5|cdot(sqrt2 - i)|=|acdot(sqrt2 - i)|=|a|cdot|sqrt2 - i|\=sqrt3 cdot a=sqrt3 cdot |2overline{z}+5|=sqrt3 cdot |overline{2z+5}|$$
answered Nov 21 at 14:23
Naweed G. Seldon
1,232419
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 at 14:31
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed G. Seldon
Nov 21 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 at 15:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
LHS is $||2overline{z}+5|cdot(sqrt2 - i)|$; and $a=|2overline{z}+5| geq 0$. Then
$$||2overline{z}+5|cdot(sqrt2 - i)|=|acdot(sqrt2 - i)|=|a|cdot|sqrt2 - i|\=sqrt3 cdot a=sqrt3 cdot |2overline{z}+5|=sqrt3 cdot |overline{2z+5}|$$
answered Nov 21 at 14:23
Naweed G. Seldon
1,232419
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 at 14:31
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed G. Seldon
Nov 21 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 at 15:03
add a comment |
up vote
1
down vote
accepted
LHS is $||2overline{z}+5|cdot(sqrt2 - i)|$; and $a=|2overline{z}+5| geq 0$. Then
$$||2overline{z}+5|cdot(sqrt2 - i)|=|acdot(sqrt2 - i)|=|a|cdot|sqrt2 - i|\=sqrt3 cdot a=sqrt3 cdot |2overline{z}+5|=sqrt3 cdot |overline{2z+5}|$$
answered Nov 21 at 14:23
Naweed G. Seldon
1,232419
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 at 14:31
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed G. Seldon
Nov 21 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 at 15:03
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
LHS is $||2overline{z}+5|cdot(sqrt2 - i)|$; and $a=|2overline{z}+5| geq 0$. Then
$$||2overline{z}+5|cdot(sqrt2 - i)|=|acdot(sqrt2 - i)|=|a|cdot|sqrt2 - i|\=sqrt3 cdot a=sqrt3 cdot |2overline{z}+5|=sqrt3 cdot |overline{2z+5}|$$
answered Nov 21 at 14:23
Naweed G. Seldon
1,232419
LHS is $||2overline{z}+5|cdot(sqrt2 - i)|$; and $a=|2overline{z}+5| geq 0$. Then
$$||2overline{z}+5|cdot(sqrt2 - i)|=|acdot(sqrt2 - i)|=|a|cdot|sqrt2 - i|\=sqrt3 cdot a=sqrt3 cdot |2overline{z}+5|=sqrt3 cdot |overline{2z+5}|$$
answered Nov 21 at 14:23
Naweed G. Seldon
1,232419
edited Nov 21 at 14:37
answered Nov 21 at 14:23
Naweed G. Seldon
1,232419
answered Nov 21 at 14:23
Naweed G. Seldon
1,232419
answered Nov 21 at 14:23
Naweed G. Seldon
1,232419
1,232419
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 at 14:31
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed G. Seldon
Nov 21 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 at 15:03
add a comment |
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 at 14:31
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed G. Seldon
Nov 21 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 at 15:03
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 at 14:31
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 at 14:31
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed G. Seldon
Nov 21 at 14:36
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed G. Seldon
Nov 21 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 at 15:03
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 at 15:03
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007745%2fshow-that-2-overlinez5-sqrt2-i-sqrt3-2z5-where-z-is-a-co%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
Nov 21 at 13:46
@PeterSzilas I didn't understand, how can you bring LHS as $sqrt3|2overline{z}+5|$ . I mean how did you proceed with the LHS to get that!?
– Kaustuv Sawarn
Nov 21 at 13:57
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 at 16:12
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 at 16:29
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 at 16:35