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 2 days ago
Kaustuv Sawarn
515
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 2 days ago
Kaustuv Sawarn
515
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
2 days ago
@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
2 days ago
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
2 days ago
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
2 days ago
Kaustuv.Welcome:)
– Peter Szilas
2 days ago
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 2 days ago
Kaustuv Sawarn
515
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 2 days ago
Kaustuv Sawarn
515
asked 2 days ago
Kaustuv Sawarn
515
asked 2 days ago
Kaustuv Sawarn
515
asked 2 days ago
Kaustuv Sawarn
515
asked 2 days ago
Kaustuv Sawarn
515
515
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
2 days ago
@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
2 days ago
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
2 days ago
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
2 days ago
Kaustuv.Welcome:)
– Peter Szilas
2 days ago
add a comment |
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
2 days ago
@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
2 days ago
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
2 days ago
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
2 days ago
Kaustuv.Welcome:)
– Peter Szilas
2 days ago
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
2 days ago
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
2 days ago
@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
2 days ago
@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
2 days ago
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
2 days ago
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
2 days ago
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
2 days ago
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
2 days ago
Kaustuv.Welcome:)
– Peter Szilas
2 days ago
Kaustuv.Welcome:)
– Peter Szilas
2 days ago
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 2 days ago
Naweed Seldon
1,212319
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
2 days ago
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed Seldon
2 days ago
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
2 days ago
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 2 days ago
Naweed Seldon
1,212319
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
2 days ago
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed Seldon
2 days ago
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
2 days ago
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 2 days ago
Naweed Seldon
1,212319
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
2 days ago
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed Seldon
2 days ago
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
2 days ago
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 2 days ago
Naweed Seldon
1,212319
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 2 days ago
Naweed Seldon
1,212319
edited 2 days ago
answered 2 days ago
Naweed Seldon
1,212319
answered 2 days ago
Naweed Seldon
1,212319
answered 2 days ago
Naweed Seldon
1,212319
1,212319
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
2 days ago
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed Seldon
2 days ago
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
2 days ago
add a comment |
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
2 days ago
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed Seldon
2 days ago
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
2 days ago
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
2 days ago
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
2 days ago
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed Seldon
2 days ago
$|a+ib| = sqrt{a^2 + b^2}$, where $z = a+ib$. In this case, $a = sqrt2, b = -1$
– Naweed Seldon
2 days ago
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
2 days ago
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
2 days ago
add a comment |
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
2 days ago
@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
2 days ago
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
2 days ago
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
2 days ago
Kaustuv.Welcome:)
– Peter Szilas
2 days ago