ACM-12 Question

[Maverick848]

I am having trouble understanding the solution to the following problem:

Find the number of ordered pairs of real numbers (a,b) such that (a +bi)^2002 = a - bi.

How do you know that z^2003=1 has 2003 solutions?

 

[soroban]

Hello, Maverick848!

I am having trouble understanding the solution to the following problem:

Find the number of ordered pairs of real numbers $\displaystyle (a,b)$ such that $\displaystyle \,(a\,+\,bi)^{2002}\:=\:a\,-\,bi$

Multiply both sides by \(\displaystyle a\,+\,bi:\;\;(a\,+\,bi)^{2003}\:=\a\,-\,bi)(a\,+\,bi)\;=\:a^2\,+\,b^2\)

Then $\displaystyle \,a\,+\,bi\;=\;$ the two thousand three $\displaystyle 2003^{rd}$ roots of $\displaystyle a^2\,+\,b^2$.

Therefore, there are 2003 solutions.

How do you know that $\displaystyle \,z^{2003}\:=\:1$ has 2003 solutions?

We have: $\displaystyle \:z\;=\;(1)^{\frac{1}{2003}}$
$\displaystyle \;\;$And every number has two square roots, three cube roots, four fourth roots, etc.

Another reason: an established theorem.
$\displaystyle \;\;\text{Every polynomial equation of degree }n\text{ has }n\text{ solutions.}$

 

[Maverick848]

The fundamental theorem of algebra guarentees that that equation has 2003 roots but it does no guarentee that they are distinct roots. Doesn't this matter for the solution?

 

[pka]

The fundamental theorem of algebra guarantees that that equation has 2003 roots but it does no guarantee that they are distinct roots. Doesn't this matter for the solution?

All of that is correct!
However, there are 2003 distinct complex 2003<SUP>rd</SUP> roots of 1.
Each of those roots is a solution of z<SUP>2003</SUP>=1.

 

[Maverick848]

But how do you know that they are distinct if the fundamental theorem does not guarentee it?

 

[Gene]

Its too late to do a search but as I recall it if you graph (a+bi)³ the answers are at 2pi/3 intervals around the unit circle. Ditto for 2pi/2003 intervals.
-----------------
Gene

 

[Matt]

But how do you know that they are distinct if the fundamental theorem does not guarentee it?

De Moivre's Theorem provides an explicit expression for all 2003 roots of unity:

\(\displaystyle \L\qquad z_k = \cos\left(\frac{2\pi k}{2003}\right)+i\sin\left(\frac{2\pi k}{2003}\right)\)

where k ranges between 0 and 2002 inclusive. It is straightfoward to check that all 2003 of these numbers are distinct.