Show that if $(z+1)^100 = (z-1)^100$, then $z$ is purely imaginary
Show that if $(z+1)^100 = (z-1)^100$, then $z$ is purely imaginary
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!
3 Answers
3
Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$
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@MohammadZuhairKhan Yes.
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– N. S.
Sep 17 '18 at 23:49
Take absolute values, $|z+1|^100=|z-1|^100$. Absolute values are always non-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 ;-) .
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is it valid to take the abs value of both sides like that?
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– Chinmayee Gidwani
Sep 18 '18 at 0:07
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Yup, if two elements are equal so are their absolute vaues.
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– Oscar Lanzi
Sep 18 '18 at 0:23
Note that $|z+1|=|z-1|$
Let $z=x+iy$
Therefore
$sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$
$(x+1)^2+y^2=(x-1)^2+y^2$
$x^2+2x+1=x^2-2x+1$
$4x=0$
$therefore x=0$
As we have found out, $z=x+iyimplies z=iy$ therefore yes, it is purely imaginary.
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So it is true for all $z in i mathbb R$?
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– CompuChip
Sep 18 '18 at 6:15
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Yes, as I just proved that there is no real component.
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– Mohammad Zuhair Khan
Sep 18 '18 at 7:42
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Umm does $overline z$ stand for the complex conjungate of $z$?
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– Mohammad Zuhair Khan
Sep 17 '18 at 23:42