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Question Number 88314 by M±th+et£s last updated on 09/Apr/20

$$(a,b)arecomplexnumbersand{a}^2+ab+b^2=0$$$$find{\left(\frac{a}{a+b}\right)}^{2020}+{\left(\frac{b}{a+b}\right)}^{2020}$$$$$$

Commented by mathmax by abdo last updated on 09/Apr/20

$$aisslutionof{x}^2+bx+b^2=0$$$$\mathrm{\Delta }=b^2-4b^2=-3b^2={\left(i\sqrt{3}b\right)}^2\Rightarrow x_1=\frac{-b+i\sqrt{3}b}{2}=b{e}^{\frac{i2\pi }{3}}$$$$and{x}_2=\frac{-b-i\sqrt{b}}{2}={be}^{-\frac{i2\pi }{3}}$$$$a=x_1\Rightarrow \frac{a}{a+b}=\frac{b{e}^{\frac{i2\pi }{3}}}{{be}^{\frac{i2\pi }{3}}+b}=\frac{e^{\frac{i2\pi }{3}}}{1+e^{\frac{i2\pi }{3}}}=\frac{1}{1+e^{-\frac{i2\pi }{3}}}=\frac{1}{1-\frac{1}{2}-\frac{i\sqrt{3}}{2}}==\frac{1}{\frac{1}{2}-\frac{i\sqrt{3}}{2}}=e^{\frac{i\pi }{3}}$$$$\frac{b}{a+b}=\frac{b}{{be}^{\frac{i2\pi }{3}}+b}=\frac{1}{e^{\frac{i2\pi }{3}}+1}=\frac{1}{1-\frac{1}{2}+\frac{i\sqrt{3}}{2}}=\frac{1}{\frac{1}{2}+\frac{i\sqrt{3}}{2}}=e^{-\frac{i\pi }{3}}\Rightarrow$$$${\left(\frac{a}{a+b}\right)}^{2020}+{\left(\frac{b}{a+b}\right)}^{2020}={\left(e^{\frac{i\pi }{3}}\right)}^{2020}+{\left(e^{-\frac{i\pi }{3}}\right)}^{2020}$$$$=2cos\left(\frac{2020\pi }{3}\right)thisvaluecanbesimplified...$$$$wrfindthesamevalueforx=x_2$$

Answered by MJS last updated on 09/Apr/20

$$a^2+ab+b^2=0$$$$\Rightarrow b=a(-\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i})$$$$\Rightarrow \frac{a}{a+b}=\frac{1}{2}\mp \frac{\sqrt{3}}{2}\mathrm{i}\wedge \frac{b}{a+b}=\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i}$$$$\Rightarrow {\left(\frac{a}{a+b}\right)}^n+{\left(\frac{b}{a+b}\right)}^n={\mathrm{e}}^{-\mathrm{i}\frac{n\pi }{3}}+{\mathrm{e}}^{\mathrm{i}\frac{n\pi }{3}}$$$$\mathrm{now}\mathrm{you}\mathrm{should}\mathrm{be}\mathrm{able}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{answer}$$

Commented by M±th+et£s last updated on 09/Apr/20

$$thanxsir.ithinkans=-1$$

Commented by MJS last updated on 09/Apr/20

$$\mathrm{yes}$$

Answered by Joel578 last updated on 10/Apr/20

$${(a+b)}^2-ab=0\Rightarrow {(a+b)}^2=ab\Rightarrow \frac{ab}{{(a+b)}^2}=1$$$$\Rightarrow \left(\frac{a}{a+b}\right)\left(\frac{b}{a+b}\right)=1$$$$$$$$\left(\frac{a}{a+b}\right)+\left(\frac{b}{a+b}\right)=\frac{a+b}{a+b}=1$$$$$$$$\mathrm{Let}x=\frac{a}{a+b},y=\frac{b}{a+b}\mathrm{with}xy=1,x+y=1$$$$\mathrm{Define}f\left(n\right)=x^n+y^n$$$$f\left(1\right)=x+y=1$$$$f\left(2\right)=x^2+y^2={(x+y)}^2-2xy=-1$$$$f\left(3\right)=x^3+y^3={(x+y)}^3-3xy(x+y)=-2$$$$f\left(4\right)=x^4+y^4=(x^3+y^3)(x+y)-xy(x^2+y^2)=-1$$$$f\left(5\right)=1,f\left(6\right)=2,f\left(7\right)=1,f\left(8\right)=-1,f\left(9\right)=-2$$$$f\left(n\right)=f(n+6)$$$$\therefore f\left(2020\right)=f\left(4\right)=-1$$

Commented by Joel578 last updated on 10/Apr/20

$$\mathrm{without}\mathrm{complex}\mathrm{method}$$

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