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I put $(x+1)^n=p(x)(x^2+1)+bx+c$ for some $p(x)$ as the other exercise where we asked to find the remainder when one polynomial is divided by another polynomial. But to make $p(x)(x^2+1)$ go so I could find $b,c$ I have to put $x=i$ which is something I shouldn't put after all. Then an idea pop into my head that the remainder itself is $(x+1)^n$, but I realized that if I put $n=2$ then the remainder is $2x$. Any idea for how to solve this?

user635988
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    Another way of doing this is to substitute $x^2=-1$ in the binomial expansion of $(x+1)^n$ and gathering the terms in $x$ (which come from odd powers of $x$ in the expansion) and constant terms (which come from even powers of $x$) – Mark Bennet Jun 30 '20 at 16:25

2 Answers2

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hint

If we replace $ x $ by $ i $ and $ -i $, we get

$$(1+i)^n=bi+c$$ $$(1-i)^n=-bi+c$$

thus $$c=\frac{(1+i)^n+(1-i)^n}{2}$$ $$b=\frac{(1+i)^n-(1-i)^n}{2i}$$

with $$1+i=\sqrt{2}e^{i\frac{\pi}{4}}$$ and $$1-i=\sqrt{2}e^{-i\frac{\pi}{4}}$$

this gives $$c=2^{\frac n2}\cos(n\frac{\pi}{4})$$ $$b=2^{\frac n2}\sin(n\frac{\pi}{4})$$

hamam_Abdallah
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Notice, $(x^2+1)$ can be factorized into linear form $(x+i)(x-i)$

According to remainder theorem, the remainder when $(x+1)^n$ is divided by $(x-i)$ is $(i+1)^n$ (i.e. obtained by substituting $x=i$ in $(x+1)^n$) which can be simplified as follows $$(1+i)^n=(\sqrt{2}e^{i\frac{\pi}{4}})^n=2^{n/2}e^{i\frac{n\pi}{4}}$$ Similarly, the remainder when $(x+1)^n$ is divided by $(x+i)$ is $(-i+1)^n$ (i.e. obtained by substituting $x=-i$ in $(x+1)^n$) which can be simplified as follows $$(1-i)^n=(\sqrt{2}e^{-i\frac{\pi}{4}})^n=2^{n/2}e^{-i\frac{n\pi}{4}}$$

Harish Chandra Rajpoot
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