Show that if $(z+1)^100 = (z-1)^100$, then $z$ is purely imaginary

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Let $z$ be a complex number satisfying
$$(z+1)^100 = (z-1)^100$$
Show that $z$ is purely imaginary, i.e. that $Re(z) = 0$.

Rearrange to

$$left(fracz+1z-1right)^100 = 1$$

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

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edited 3 mins ago

TheSimpliFire

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up vote
7
down vote

favorite

Let $z$ be a complex number satisfying
$$(z+1)^100 = (z-1)^100$$
Show that $z$ is purely imaginary, i.e. that $Re(z) = 0$.

Rearrange to

$$left(fracz+1z-1right)^100 = 1$$

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

share|cite|improve this question

edited 3 mins ago

TheSimpliFire

11k62154

asked 7 hours ago

Chinmayee Gidwani

462

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 | 

up vote
7
down vote

favorite

up vote
7
down vote

favorite

Let $z$ be a complex number satisfying
$$(z+1)^100 = (z-1)^100$$
Show that $z$ is purely imaginary, i.e. that $Re(z) = 0$.

Rearrange to

$$left(fracz+1z-1right)^100 = 1$$

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

share|cite|improve this question

edited 3 mins ago

TheSimpliFire

11k62154

asked 7 hours ago

Chinmayee Gidwani

462

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.

Let $z$ be a complex number satisfying
$$(z+1)^100 = (z-1)^100$$
Show that $z$ is purely imaginary, i.e. that $Re(z) = 0$.

Rearrange to

$$left(fracz+1z-1right)^100 = 1$$

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

complex-analysis complex-numbers

share|cite|improve this question

edited 3 mins ago

TheSimpliFire

11k62154

asked 7 hours ago

Chinmayee Gidwani

462

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.

share|cite|improve this question

edited 3 mins ago

TheSimpliFire

11k62154

asked 7 hours ago

Chinmayee Gidwani

462

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.

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share|cite|improve this question

edited 3 mins ago

TheSimpliFire

11k62154

edited 3 mins ago

TheSimpliFire

11k62154

edited 3 mins ago

TheSimpliFire

11k62154

11k62154

asked 7 hours ago

Chinmayee Gidwani

462

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 7 hours ago

Chinmayee Gidwani

462

asked 7 hours ago

Chinmayee Gidwani

462

462

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 | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
10
down vote



accepted

Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$

share|cite|improve this answer

answered 7 hours ago

N. S.

98.8k5106197

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Umm does $overline z$ stand for the complex conjungate of $z$?
  – Mohammad Zuhair Khan
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MohammadZuhairKhan Yes.
  – N. S.
  7 hours ago
  

  
  
  
  

add a comment | 

up vote
4
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 ;-) .

share|cite|improve this answer

answered 7 hours ago

Oscar Lanzi

10.4k11733

• 
  
  
  

  
  

  
  
  
  
  
  

  
  is it valid to take the abs value of both sides like that?
  – Chinmayee Gidwani
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yup, if two elements are equal so are their absolute vaues.
  – Oscar Lanzi
  6 hours ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

Note that $|z+1|=|z-1|$

Therefore

$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$

$(x+1)^2+y^2=(x-1)^2+y^2$

$x^2+2x+1=x^2-2x+1$

$4x=0$

$therefore x=0$

share|cite|improve this answer

answered 7 hours ago

Mohammad Zuhair Khan

938422

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So it is true for all $z in i mathbb R$?
  – CompuChip
  39 mins ago
  

  
  
  
  

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
10
down vote



accepted

Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$

share|cite|improve this answer

answered 7 hours ago

N. S.

98.8k5106197

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Umm does $overline z$ stand for the complex conjungate of $z$?
  – Mohammad Zuhair Khan
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MohammadZuhairKhan Yes.
  – N. S.
  7 hours ago
  

  
  
  
  

add a comment | 

up vote
10
down vote



accepted

Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$

share|cite|improve this answer

answered 7 hours ago

N. S.

98.8k5106197

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Umm does $overline z$ stand for the complex conjungate of $z$?
  – Mohammad Zuhair Khan
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MohammadZuhairKhan Yes.
  – N. S.
  7 hours ago
  

  
  
  
  

add a comment | 

up vote
10
down vote



accepted

up vote
10
down vote



accepted

Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$

share|cite|improve this answer

answered 7 hours ago

N. S.

98.8k5106197

Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$

share|cite|improve this answer

answered 7 hours ago

N. S.

98.8k5106197

share|cite|improve this answer

share|cite|improve this answer

answered 7 hours ago

N. S.

98.8k5106197

answered 7 hours ago

N. S.

98.8k5106197

answered 7 hours ago

N. S.

98.8k5106197

98.8k5106197

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Umm does $overline z$ stand for the complex conjungate of $z$?
  – Mohammad Zuhair Khan
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MohammadZuhairKhan Yes.
  – N. S.
  7 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Umm does $overline z$ stand for the complex conjungate of $z$?
  – Mohammad Zuhair Khan
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MohammadZuhairKhan Yes.
  – N. S.
  7 hours ago
  

  
  
  
  

Umm does $overline z$ stand for the complex conjungate of $z$?
– Mohammad Zuhair Khan
7 hours ago

Umm does $overline z$ stand for the complex conjungate of $z$?
– Mohammad Zuhair Khan
7 hours ago

@MohammadZuhairKhan Yes.
– N. S.
7 hours ago

@MohammadZuhairKhan Yes.
– N. S.
7 hours ago

add a comment | 

up vote
4
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 ;-) .

share|cite|improve this answer

answered 7 hours ago

Oscar Lanzi

10.4k11733

• 
  
  
  

  
  

  
  
  
  
  
  

  
  is it valid to take the abs value of both sides like that?
  – Chinmayee Gidwani
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yup, if two elements are equal so are their absolute vaues.
  – Oscar Lanzi
  6 hours ago
  

  
  
  
  

add a comment | 

up vote
4
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 ;-) .

share|cite|improve this answer

answered 7 hours ago

Oscar Lanzi

10.4k11733

• 
  
  
  

  
  

  
  
  
  
  
  

  
  is it valid to take the abs value of both sides like that?
  – Chinmayee Gidwani
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yup, if two elements are equal so are their absolute vaues.
  – Oscar Lanzi
  6 hours ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
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 ;-) .

share|cite|improve this answer

answered 7 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 ;-) .

share|cite|improve this answer

answered 7 hours ago

Oscar Lanzi

10.4k11733

share|cite|improve this answer

share|cite|improve this answer

answered 7 hours ago

Oscar Lanzi

10.4k11733

answered 7 hours ago

Oscar Lanzi

10.4k11733

answered 7 hours ago

Oscar Lanzi

10.4k11733

10.4k11733

• 
  
  
  

  
  

  
  
  
  
  
  

  
  is it valid to take the abs value of both sides like that?
  – Chinmayee Gidwani
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yup, if two elements are equal so are their absolute vaues.
  – Oscar Lanzi
  6 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  is it valid to take the abs value of both sides like that?
  – Chinmayee Gidwani
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yup, if two elements are equal so are their absolute vaues.
  – Oscar Lanzi
  6 hours ago
  

  
  
  
  

is it valid to take the abs value of both sides like that?
– Chinmayee Gidwani
6 hours ago

is it valid to take the abs value of both sides like that?
– Chinmayee Gidwani
6 hours ago

Yup, if two elements are equal so are their absolute vaues.
– Oscar Lanzi
6 hours ago

Yup, if two elements are equal so are their absolute vaues.
– Oscar Lanzi
6 hours ago

add a comment | 

up vote
3
down vote

Note that $|z+1|=|z-1|$

Therefore

$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$

$(x+1)^2+y^2=(x-1)^2+y^2$

$x^2+2x+1=x^2-2x+1$

$4x=0$

$therefore x=0$

share|cite|improve this answer

answered 7 hours ago

Mohammad Zuhair Khan

938422

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So it is true for all $z in i mathbb R$?
  – CompuChip
  39 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

Note that $|z+1|=|z-1|$

Therefore

$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$

$(x+1)^2+y^2=(x-1)^2+y^2$

$x^2+2x+1=x^2-2x+1$

$4x=0$

$therefore x=0$

share|cite|improve this answer

answered 7 hours ago

Mohammad Zuhair Khan

938422

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So it is true for all $z in i mathbb R$?
  – CompuChip
  39 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Note that $|z+1|=|z-1|$

Therefore

$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$

$(x+1)^2+y^2=(x-1)^2+y^2$

$x^2+2x+1=x^2-2x+1$

$4x=0$

$therefore x=0$

share|cite|improve this answer

answered 7 hours ago

Mohammad Zuhair Khan

938422

Note that $|z+1|=|z-1|$

Therefore

$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$

$(x+1)^2+y^2=(x-1)^2+y^2$

$x^2+2x+1=x^2-2x+1$

$4x=0$

$therefore x=0$

share|cite|improve this answer

answered 7 hours ago

Mohammad Zuhair Khan

938422

share|cite|improve this answer

share|cite|improve this answer

answered 7 hours ago

Mohammad Zuhair Khan

938422

answered 7 hours ago

Mohammad Zuhair Khan

938422

answered 7 hours ago

Mohammad Zuhair Khan

938422

938422

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So it is true for all $z in i mathbb R$?
  – CompuChip
  39 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So it is true for all $z in i mathbb R$?
  – CompuChip
  39 mins ago
  

  
  
  
  

So it is true for all $z in i mathbb R$?
– CompuChip
39 mins ago

So it is true for all $z in i mathbb R$?
– CompuChip
39 mins ago

add a comment | 

Chinmayee Gidwani is a new contributor. Be nice, and check out our Code of Conduct.

 

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