Qs involving a/b+b/a+1=0 COMPLEX NUMBERS (1 Viewer)

gamja · Jan 13, 2023

Let a, b be complex numbers that satisfy $\frac{a}{b}+\frac{b}{a}+1=0.$

i) Prove that $\frac{a^{3n}}{b^{3n}}+\frac{b^{3n}}{a^{3n}}=2$ for n≥1, by using Mathematical Induction.
- I proved true for n=1 and assumed true for n=k, just need help with Proving true for n=k+1

ii) Show that $\frac{a^n}{b^n}+\frac{b^n}{a^n}=-1$ , if n is not a multiple of 3.

Thanks in advance for help!

tywebb · Jan 14, 2023

i. So you proved it true for n=1 in which case you know that $\frac{a^3}{b^3}+\frac{b^3}{a^3}=2$ and hence $(a^3-b^3)^2=0$ so $a^3=b^3$

Likewise for induction hypothesis, if true for n=k then $\frac{a^{3k}}{b^{3k}}+\frac{b^{3k}}{a^{3k}}=2$ and hence $(a^{3k}-b^{3k})^2=0$ so $a^{3k}=b^{3k}$ .

Then for n=k+1 we have $\frac{a^{3k+3}}{b^{3k+3}}+\frac{b^{3k+3}}{a^{3k+3}}=\frac{a^3}{b^3}+\frac{b^3}{a^3}=2$

ii.

Case 1, n=3k+1

$\frac{a^{3k+1}}{b^{3k+1}}+\frac{b^{3k+1}}{a^{3k+1}}=\frac{a}{b}+\frac{b}{a}=-1$

Case 2, n=3k+2 ( a little trickier)

$\frac{a^{3k+2}}{b^{3k+2}}+\frac{b^{3k+2}}{a^{3k+2}}=\frac{a^2}{b^2}+\frac{b^2}{a^2}=(\frac{a}{b}+\frac{b}{a})^2-2\frac{a}{b}\frac{b}{a}=(-1)^2-2=-1$

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Note: Although $a^3=b^3$ we can't say $a=b$ because then $\frac{a}{b}+\frac{b}{a}+1=3$ , not 0.

However we can say $a=\omega b$ or $\omega^2b$ where $\omega$ is a complex cube root of unity.

Then we have $\frac{a}{b}+\frac{b}{a}+1=\omega+\frac{1}{\omega}+1=\omega+\omega^2+1=0$ as expected or

$\frac{a}{b}+\frac{b}{a}+1=\omega^2+\frac{1}{\omega^2}+1=\omega^2+\omega+1=0$ as expected.

Qs involving a/b+b/a+1=0 COMPLEX NUMBERS (1 Viewer)

gamja

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tywebb

dangerman

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