Mixing
Show that z is purely imaginary
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Show that z is a purely imaginary number given that
$(z+1)^100 = (z-1)^100$
Rearrange to
$(fracz+1z-1)^100 = 1$
I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^ipi$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!
complex-analysis complex-numbers proof-writing
edited 2 hours ago
abc...
2,134529
asked 2 hours ago
Chinmayee Gidwani
212
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Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
2
down vote
favorite
Show that z is a purely imaginary number given that
$(z+1)^100 = (z-1)^100$
Rearrange to
$(fracz+1z-1)^100 = 1$
I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^ipi$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!
complex-analysis complex-numbers proof-writing
edited 2 hours ago
abc...
2,134529
asked 2 hours ago
Chinmayee Gidwani
212
New contributor
Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Show that z is a purely imaginary number given that
$(z+1)^100 = (z-1)^100$
Rearrange to
$(fracz+1z-1)^100 = 1$
I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^ipi$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!
complex-analysis complex-numbers proof-writing
edited 2 hours ago
abc...
2,134529
asked 2 hours ago
Chinmayee Gidwani
212
New contributor
Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Show that z is a purely imaginary number given that
$(z+1)^100 = (z-1)^100$
Rearrange to
$(fracz+1z-1)^100 = 1$
I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^ipi$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!
complex-analysis complex-numbers proof-writing
complex-analysis complex-numbers proof-writing
edited 2 hours ago
abc...
2,134529
asked 2 hours ago
Chinmayee Gidwani
212
New contributor
Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
abc...
2,134529
asked 2 hours ago
Chinmayee Gidwani
212
New contributor
Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
abc...
2,134529
edited 2 hours ago
abc...
2,134529
edited 2 hours ago
abc...
2,134529
2,134529
asked 2 hours ago
Chinmayee Gidwani
212
New contributor
Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Chinmayee Gidwani
212
asked 2 hours ago
Chinmayee Gidwani
212
212
New contributor
Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$
answered 2 hours ago
N. S.
98.8k5106197
Umm does $overline z$ stand for the complex conjungate of $z$?
– Mohammad Zuhair Khan
2 hours ago
@MohammadZuhairKhan Yes.
– N. S.
2 hours ago
add a comment |Â
up vote
1
down vote
Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^th$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .
answered 2 hours ago
Oscar Lanzi
10.4k11733
is it valid to take the abs value of both sides like that?
– Chinmayee Gidwani
1 hour ago
Yup, if two elements are equal so are their absolute vaues.
– Oscar Lanzi
1 hour ago
add a comment |Â
up vote
1
down vote
Note that $|z+1|=|z-1|$
Therefore
$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$
$(x+1)^2+y^2=(x-1)^2+y^2$
$x^2+2x+1=x^2-2x+1$
$4x=0$
$therefore x=0$
answered 2 hours ago
Mohammad Zuhair Khan
918422
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$
answered 2 hours ago
N. S.
98.8k5106197
Umm does $overline z$ stand for the complex conjungate of $z$?
– Mohammad Zuhair Khan
2 hours ago
@MohammadZuhairKhan Yes.
– N. S.
2 hours ago
add a comment |Â
up vote
4
down vote
accepted
Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$
answered 2 hours ago
N. S.
98.8k5106197
Umm does $overline z$ stand for the complex conjungate of $z$?
– Mohammad Zuhair Khan
2 hours ago
@MohammadZuhairKhan Yes.
– N. S.
2 hours ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$
answered 2 hours ago
N. S.
98.8k5106197
Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$
answered 2 hours ago
N. S.
98.8k5106197
answered 2 hours ago
N. S.
98.8k5106197
answered 2 hours ago
N. S.
98.8k5106197
answered 2 hours ago
N. S.
98.8k5106197
98.8k5106197
Umm does $overline z$ stand for the complex conjungate of $z$?
– Mohammad Zuhair Khan
2 hours ago
@MohammadZuhairKhan Yes.
– N. S.
2 hours ago
add a comment |Â
Umm does $overline z$ stand for the complex conjungate of $z$?
– Mohammad Zuhair Khan
2 hours ago
@MohammadZuhairKhan Yes.
– N. S.
2 hours ago
Umm does $overline z$ stand for the complex conjungate of $z$?
– Mohammad Zuhair Khan
2 hours ago
Umm does $overline z$ stand for the complex conjungate of $z$?
– Mohammad Zuhair Khan
2 hours ago
@MohammadZuhairKhan Yes.
– N. S.
2 hours ago
@MohammadZuhairKhan Yes.
– N. S.
2 hours ago
add a comment |Â
up vote
1
down vote
Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^th$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .
answered 2 hours ago
Oscar Lanzi
10.4k11733
is it valid to take the abs value of both sides like that?
– Chinmayee Gidwani
1 hour ago
Yup, if two elements are equal so are their absolute vaues.
– Oscar Lanzi
1 hour ago
add a comment |Â
up vote
1
down vote
Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^th$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .
answered 2 hours ago
Oscar Lanzi
10.4k11733
is it valid to take the abs value of both sides like that?
– Chinmayee Gidwani
1 hour ago
Yup, if two elements are equal so are their absolute vaues.
– Oscar Lanzi
1 hour ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^th$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .
answered 2 hours ago
Oscar Lanzi
10.4k11733
Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^th$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .
answered 2 hours ago
Oscar Lanzi
10.4k11733
answered 2 hours ago
Oscar Lanzi
10.4k11733
answered 2 hours ago
Oscar Lanzi
10.4k11733
answered 2 hours ago
Oscar Lanzi
10.4k11733
10.4k11733
is it valid to take the abs value of both sides like that?
– Chinmayee Gidwani
1 hour ago
Yup, if two elements are equal so are their absolute vaues.
– Oscar Lanzi
1 hour ago
add a comment |Â
is it valid to take the abs value of both sides like that?
– Chinmayee Gidwani
1 hour ago
Yup, if two elements are equal so are their absolute vaues.
– Oscar Lanzi
1 hour ago
is it valid to take the abs value of both sides like that?
– Chinmayee Gidwani
1 hour ago
is it valid to take the abs value of both sides like that?
– Chinmayee Gidwani
1 hour ago
Yup, if two elements are equal so are their absolute vaues.
– Oscar Lanzi
1 hour ago
Yup, if two elements are equal so are their absolute vaues.
– Oscar Lanzi
1 hour ago
add a comment |Â
up vote
1
down vote
Note that $|z+1|=|z-1|$
Therefore
$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$
$(x+1)^2+y^2=(x-1)^2+y^2$
$x^2+2x+1=x^2-2x+1$
$4x=0$
$therefore x=0$
answered 2 hours ago
Mohammad Zuhair Khan
918422
add a comment |Â
up vote
1
down vote
Note that $|z+1|=|z-1|$
Therefore
$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$
$(x+1)^2+y^2=(x-1)^2+y^2$
$x^2+2x+1=x^2-2x+1$
$4x=0$
$therefore x=0$
answered 2 hours ago
Mohammad Zuhair Khan
918422
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that $|z+1|=|z-1|$
Therefore
$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$
$(x+1)^2+y^2=(x-1)^2+y^2$
$x^2+2x+1=x^2-2x+1$
$4x=0$
$therefore x=0$
answered 2 hours ago
Mohammad Zuhair Khan
918422
Note that $|z+1|=|z-1|$
Therefore
$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$
$(x+1)^2+y^2=(x-1)^2+y^2$
$x^2+2x+1=x^2-2x+1$
$4x=0$
$therefore x=0$
answered 2 hours ago
Mohammad Zuhair Khan
918422
answered 2 hours ago
Mohammad Zuhair Khan
918422
answered 2 hours ago
Mohammad Zuhair Khan
918422
answered 2 hours ago
Mohammad Zuhair Khan
918422
918422
add a comment |Â
add a comment |Â
Chinmayee Gidwani is a new contributor. Be nice, and check out our Code of Conduct.
Chinmayee Gidwani is a new contributor. Be nice, and check out our Code of Conduct.
Chinmayee Gidwani is a new contributor. Be nice, and check out our Code of Conduct.
Chinmayee Gidwani is a new contributor. Be nice, and check out our Code of Conduct.
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