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How to find the value of $x$ in $(x^{2}+1)^{35}=0$? we can find value of $x$ in $(x^{2}+1)^{2}=0$ by using algebraic identity i.e. $$x^{2}+2x+1=0$$ $$x^{2}+x+x+1=0$$ $$x(x+1)+1(x+1)=0$$ $$(x+1)(x+1)=0$$ $$x=-1$$ but how do we find value of $x$ in $(x^{2}+1)^{35}=0$?

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1 Answer

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There is no real root of $(x^2+1)^{35}=0$, because $$ (x^2+1)^{35}\ge 1^{35}=1>0. $$ Now, find roots in $\mathbb{C}$. Since $x^2+1=(x+i)(x-i)$, \begin{align} (x^2+1)^{35}=(x-i)^{35}(x+i)^{35}=0. \end{align} Thus $(x-i)^{35}=0$ or $(x+i)^{35}=0$. Therefore, $x-i=0$ or $x+i=0$. $$ \therefore x=\pm i $$

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