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Question Number 128574 by mohammad17 last updated on 08/Jan/21

Does the function f(z)=((3z^4 −2z^3 +8z^2 −2z+5)/(z−i))     continuous at z=i ?

$${Does}\:{the}\:{function}\:{f}\left({z}\right)=\frac{\mathrm{3}{z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{3}} +\mathrm{8}{z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{5}}{{z}−{i}}\: \\ $$$$ \\ $$$${continuous}\:{at}\:{z}={i}\:? \\ $$

Answered by MJS_new last updated on 08/Jan/21

f(z)=(((z^2 +1)(3z^2 −2z+5))/(z−i))=  =(((z−i)(z+i)(3z^2 −2z+5))/(z−i))=  =(z+i)(3z^2 −2z+5)  ⇒ f(z) is continuous at z=i

$${f}\left({z}\right)=\frac{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{5}\right)}{{z}−\mathrm{i}}= \\ $$$$=\frac{\left({z}−\mathrm{i}\right)\left({z}+\mathrm{i}\right)\left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{5}\right)}{{z}−\mathrm{i}}= \\ $$$$=\left({z}+\mathrm{i}\right)\left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{5}\right) \\ $$$$\Rightarrow\:{f}\left({z}\right)\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{at}\:{z}=\mathrm{i} \\ $$

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