Complex Number Q (1 Viewer)

[Vampire]

Hey guys, I think this is a pretty generic sort of question but I couldn't figure it out...any ideas?

Show that if z satisfies the equation (z-i)^13 = (z+i)^13, then z must be real.

Thx

 

[Riviet]

Just an idea, but maybe you could prove it by contradiction. We would assume that z satisifies the equation when z is imaginary. Let z=ai, where "a" is real.

Then, (ai-i)¹³=(ai+i)¹³
[i(a-1)]¹³=[i(a+1)]¹³
i(a-1)¹³=i(a+1)¹³, since i¹³=i
a-1=a+1
.'. no solutions when z=ai, where "a" is real.

Hence "a" must be imaginary, meaning z is real.

Just having a stab at it... does that work?

 

[Gowr]

hmmm... that is tentative at best considering that you haven't considered the case where z=a+ib

It seems that there is a much simpler solution....

as (z+i)^13=(z-i)^13
|(z+i)^13| = |(z-i))^13|

Just take the 13th roots of both sides (no reason you cant do that), and get |z+i| = |z-i|, the solution of that being quite obviously any point on the real axis

 

[Vampire]

Hmmm...I'd like to see icycloud's opinion on this too if you're there...thx

 

[Gowr]

My opinion isn't valued... :burn:

But quite seriously, this method does work. The equivalence of modulii is a perfectly sound method. I mean, consider this on the argand plane. If (z+i)^13=(z-i)^13, then z+i and z-i obviously have to have the same modulus, or if you take them to the power of 13 they will quite clearly be different, and for them to have the same modulus they obviously must lie on the perpendicular bisector of the segment joining (0,1) and (0,-1), that is the real axis.

It can be seen that, as expected, there are 13 solutions to this, that is where 1/z = tan n*pi/13 for n= 0 to 12 noting that z is real, which is what you would expect for a polynomial of degree 13

 

[Riviet]

Just take the 13th roots of both sides (no reason you cant do that), and get |z+i| = |z-i|, the solution of that being quite obviously any point on the real axis

Very clever... I like that method.

 

[Rax]

Very clever... I like that method.

Yep, that method makes perfect sense

Keep up the nice Logical work

 

[icycloud]

Hmmm...I'd like to see icycloud's opinion on this too if you're there...thx

Wow I'm so flustered *blushes* Many of the people who have replied to this thread are the same level or better than me at Maths, so you can safely trust their answers But anyway I agree that Gowr's method does work and is a good way for doing this question.