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Question-94193




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
thank you sir but a=−(1/(2i))
$${thank}\:{you}\:{sir}\:{but}\:{a}=−\frac{\mathrm{1}}{\mathrm{2}{i}} \\ $$
Commented by prakash jain last updated on 17/May/20
Thank You.  corrected
$$\mathrm{Thank}\:\mathrm{You}. \\ $$$$\mathrm{corrected} \\ $$
Commented by prakash jain last updated on 17/May/20
(z+i)(z−i)=z^2 +1  Let us say that remainer is az+b  as z^2  is of degree 2 remainder will  be of degree 1.  f(z)=g(z)(z^2 +1)+(az+b)   (I)  remainder when f(z) is divided  by z+i is given by f(−i).  we put z=−i in I  f(−i)=−ai+b=1+i    (II)  Similarly for z−i  f(i)=ai+b=i                   (III)  Solving for a and b from II and III.  b=((1+2i)/2),  a=−(1/(2i))  remainer when f(z) is divided  by z^2 +1 is (az+b) where a and b  are given above.
$$\left({z}+{i}\right)\left({z}−{i}\right)={z}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:\mathrm{that}\:\mathrm{remainer}\:\mathrm{is}\:{az}+{b} \\ $$$$\mathrm{as}\:{z}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{2}\:\mathrm{remainder}\:\mathrm{will} \\ $$$$\mathrm{be}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{1}. \\ $$$${f}\left({z}\right)={g}\left({z}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)+\left({az}+{b}\right)\:\:\:\left(\mathrm{I}\right) \\ $$$$\mathrm{remainder}\:\mathrm{when}\:{f}\left({z}\right)\:\mathrm{is}\:\mathrm{divided} \\ $$$$\mathrm{by}\:{z}+{i}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:{f}\left(−{i}\right). \\ $$$$\mathrm{we}\:\mathrm{put}\:{z}=−{i}\:\mathrm{in}\:\mathrm{I} \\ $$$${f}\left(−{i}\right)=−{ai}+{b}=\mathrm{1}+{i}\:\:\:\:\left(\mathrm{II}\right) \\ $$$$\mathrm{Similarly}\:\mathrm{for}\:{z}−{i} \\ $$$${f}\left({i}\right)={ai}+{b}={i}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{III}\right) \\ $$$$\mathrm{Solving}\:\mathrm{for}\:{a}\:\mathrm{and}\:{b}\:\mathrm{from}\:\mathrm{II}\:\mathrm{and}\:\mathrm{III}. \\ $$$${b}=\frac{\mathrm{1}+\mathrm{2}{i}}{\mathrm{2}}, \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{2}{i}} \\ $$$${r}\mathrm{emainer}\:\mathrm{when}\:{f}\left({z}\right)\:\mathrm{is}\:\mathrm{divided} \\ $$$$\mathrm{by}\:{z}^{\mathrm{2}} +\mathrm{1}\:\mathrm{is}\:\left({az}+{b}\right)\:\mathrm{where}\:{a}\:\mathrm{and}\:{b} \\ $$$$\mathrm{are}\:\mathrm{given}\:\mathrm{above}. \\ $$

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