Mixing
Show that $|{|2overline{z}+5 |(sqrt2 - i)} | = sqrt3 |2z+5|$, where z is a complex number.
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 '18 at 13:40
Kaustuv Sawarn
465
add a comment |
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 '18 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 '18 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 '18 at 13:57
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 '18 at 16:12
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 '18 at 16:29
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 '18 at 16:35
add a comment |
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 '18 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 '18 at 13:40
Kaustuv Sawarn
465
asked Nov 21 '18 at 13:40
Kaustuv Sawarn
465
asked Nov 21 '18 at 13:40
Kaustuv Sawarn
465
asked Nov 21 '18 at 13:40
Kaustuv Sawarn
465
asked Nov 21 '18 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 '18 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 '18 at 13:57
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 '18 at 16:12
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 '18 at 16:29
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 '18 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 '18 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 '18 at 13:57
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 '18 at 16:12
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 '18 at 16:29
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 '18 at 16:35
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
Nov 21 '18 at 13:46
|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
Nov 21 '18 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 '18 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 '18 at 13:57
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 '18 at 16:12
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 '18 at 16:12
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 '18 at 16:29
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 '18 at 16:29
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 '18 at 16:35
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 '18 at 16:35
add a comment |
1 Answer
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oldest
votes
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 '18 at 14:23
Naweed G. Seldon
1,309419
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 '18 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 '18 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 '18 at 15:03
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
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active
oldest
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votes
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 '18 at 14:23
Naweed G. Seldon
1,309419
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 '18 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 '18 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 '18 at 15:03
add a comment |
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 '18 at 14:23
Naweed G. Seldon
1,309419
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 '18 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 '18 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 '18 at 15:03
add a comment |
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 '18 at 14:23
Naweed G. Seldon
1,309419
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 '18 at 14:23
Naweed G. Seldon
1,309419
edited Nov 21 '18 at 14:37
answered Nov 21 '18 at 14:23
Naweed G. Seldon
1,309419
answered Nov 21 '18 at 14:23
Naweed G. Seldon
1,309419
answered Nov 21 '18 at 14:23
Naweed G. Seldon
1,309419
1,309419
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 '18 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 '18 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 '18 at 15:03
add a comment |
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 '18 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 '18 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 '18 at 15:03
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 '18 at 14:31
I think this is a silly question but how is $|sqrt2-i| = sqrt3$
– Kaustuv Sawarn
Nov 21 '18 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 '18 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 '18 at 14:36
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 '18 at 15:03
haha, damn. I didn't even think like that! Thankyou very much!!
– Kaustuv Sawarn
Nov 21 '18 at 15:03
add a comment |
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|ab|= |a||b|, a,b complex.LHS: $√3|2overline{z}+5|$.Then $|a|=|overline{a}|$.Helps?
– Peter Szilas
Nov 21 '18 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 '18 at 13:57
Kaustuv.$|√2-i|=sqrt{(√2)^2+i^2}=sqrt{4-1} =√3.$Ok?
– Peter Szilas
Nov 21 '18 at 16:12
@PeterSzilas yes. I got it. Thanks!
– Kaustuv Sawarn
Nov 21 '18 at 16:29
Kaustuv.Welcome:)
– Peter Szilas
Nov 21 '18 at 16:35