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Question Number 94193 by seedhamaieng@gmail.com last updated on 17/May/20

Commented by seedhamaieng@gmail.com last updated on 17/May/20

translate into Hind language Question no 7

Commented by seedhamaieng@gmail.com last updated on 17/May/20

Commented by seedhamaieng@gmail.com last updated on 17/May/20

$$thankyousirbuta=-\frac{1}{2i}$$

Commented by prakash jain last updated on 17/May/20

$$\mathrm{Thank}\mathrm{You}.$$$$\mathrm{corrected}$$

Commented by prakash jain last updated on 17/May/20

$$(z+i)(z-i)=z^2+1$$$$\mathrm{Let}\mathrm{us}\mathrm{say}\mathrm{that}\mathrm{remainer}\mathrm{is}az+b$$$$\mathrm{as}{z}^2\mathrm{is}\mathrm{of}\mathrm{degree}2\mathrm{remainder}\mathrm{will}$$$$\mathrm{be}\mathrm{of}\mathrm{degree}1.$$$$f\left(z\right)=g\left(z\right)(z^2+1)+(az+b)\left(\mathrm{I}\right)$$$$\mathrm{remainder}\mathrm{when}f\left(z\right)\mathrm{is}\mathrm{divided}$$$$\mathrm{by}z+i\mathrm{is}\mathrm{given}\mathrm{by}f(-i).$$$$\mathrm{we}\mathrm{put}z=-i\mathrm{in}\mathrm{I}$$$$f(-i)=-ai+b=1+i\left(\mathrm{II}\right)$$$$\mathrm{Similarly}\mathrm{for}z-i$$$$f\left(i\right)=ai+b=i\left(\mathrm{III}\right)$$$$\mathrm{Solving}\mathrm{for}a\mathrm{and}b\mathrm{from}\mathrm{II}\mathrm{and}\mathrm{III}.$$$$b=\frac{1+2i}{2},$$$$a=-\frac{1}{2i}$$$$r\mathrm{emainer}\mathrm{when}f\left(z\right)\mathrm{is}\mathrm{divided}$$$$\mathrm{by}{z}^2+1\mathrm{is}(az+b)\mathrm{where}a\mathrm{and}b$$$$\mathrm{are}\mathrm{given}\mathrm{above}.$$

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