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Question-126827

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Question Number 126827 by harckinwunmy last updated on 24/Dec/20
Answered by AST last updated on 24/Dec/20
Q33. ((4+7i)/(1−3i)) Rationalising ⇒ (((4+7i)(1+3i))/((1−3i)(1+3i)))=((4+12i+7i+(21i^2 ))/(1^2 −(3i)^2 )) =((4+19i−21)/(1−(−9)))=((−17+19i)/(10))=((−17)/(10))+i(((19)/(10))) =−1.7+1.9i (C)
$$\mathrm{Q33}. \\ $$$$\frac{\mathrm{4}+\mathrm{7}{i}}{\mathrm{1}−\mathrm{3}{i}} \\ $$$$\mathrm{Rationalising} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{4}+\mathrm{7}{i}\right)\left(\mathrm{1}+\mathrm{3}{i}\right)}{\left(\mathrm{1}−\mathrm{3}{i}\right)\left(\mathrm{1}+\mathrm{3}{i}\right)}=\frac{\mathrm{4}+\mathrm{12}{i}+\mathrm{7}{i}+\left(\mathrm{21}{i}^{\mathrm{2}} \right)}{\mathrm{1}^{\mathrm{2}} −\left(\mathrm{3}{i}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}+\mathrm{19}{i}−\mathrm{21}}{\mathrm{1}−\left(−\mathrm{9}\right)}=\frac{−\mathrm{17}+\mathrm{19}{i}}{\mathrm{10}}=\frac{−\mathrm{17}}{\mathrm{10}}+{i}\left(\frac{\mathrm{19}}{\mathrm{10}}\right) \\ $$$$=−\mathrm{1}.\mathrm{7}+\mathrm{1}.\mathrm{9}{i}\:\left({C}\right) \\ $$
Answered by AST last updated on 24/Dec/20
Q34. Rationalising ((1+i)/(2+i)) ⇒(((1+i)(2−i))/((2+i)(2−i)))=((2−i+2i−i^2 )/(4−i^2 ))=((2+i−(−1))/(4−(−1))) =((3+i)/5)=(3/5)+(1/5)i ⇒x=(3/5)
$$\mathrm{Q34}. \\ $$$$\mathrm{Rationalising}\:\frac{\mathrm{1}+{i}}{\mathrm{2}+{i}} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}+{i}\right)\left(\mathrm{2}−{i}\right)}{\left(\mathrm{2}+{i}\right)\left(\mathrm{2}−{i}\right)}=\frac{\mathrm{2}−{i}+\mathrm{2}{i}−{i}^{\mathrm{2}} }{\mathrm{4}−{i}^{\mathrm{2}} }=\frac{\mathrm{2}+{i}−\left(−\mathrm{1}\right)}{\mathrm{4}−\left(−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{3}+{i}}{\mathrm{5}}=\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}{i} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$
Answered by AST last updated on 24/Dec/20
Q35. Rationalising z ⇒ ((1(1−2i))/((1+2i)(1−2i)))=((1−2i)/(1−(2i)^2 ))=((1−2i)/(1−(−4))) =((1−2i)/5)=(1/5)−(2/5)i z^− =(1/5)+(2/5)i
$$\mathrm{Q35}. \\ $$$$\mathrm{Rationalising}\:\mathrm{z} \\ $$$$\Rightarrow\:\frac{\mathrm{1}\left(\mathrm{1}−\mathrm{2}{i}\right)}{\left(\mathrm{1}+\mathrm{2}{i}\right)\left(\mathrm{1}−\mathrm{2}{i}\right)}=\frac{\mathrm{1}−\mathrm{2}{i}}{\mathrm{1}−\left(\mathrm{2}{i}\right)^{\mathrm{2}} }=\frac{\mathrm{1}−\mathrm{2}{i}}{\mathrm{1}−\left(−\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{1}−\mathrm{2}{i}}{\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{5}}{i} \\ $$$$\overset{−} {{z}}=\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{5}}{i} \\ $$
Answered by AST last updated on 24/Dec/20
Q36. Note that z^�_− = z ⇒ conjugate of (conjugate of z) = z Hence, answer is B.
$$\mathrm{Q36}. \\ $$$$\mathrm{Note}\:\mathrm{that}\:\bar {\mathrm{z}}\:=\:\mathrm{z} \\ $$$$\Rightarrow\:\mathrm{conjugate}\:\mathrm{of}\:\left(\mathrm{conjugate}\:\mathrm{of}\:\mathrm{z}\right)\:=\:\mathrm{z} \\ $$$$\mathrm{Hence},\:\mathrm{answer}\:\mathrm{is}\:\mathrm{B}. \\ $$
Commented by harckinwunmy last updated on 24/Dec/20
thanks AST. can you help complete it
$${thanks}\:{AST}.\:{can}\:{you}\:{help}\:{complete}\:{it} \\ $$
Answered by AST last updated on 24/Dec/20
Q37. Rationalising z ⇒ ((1(1+i))/((1−i)(1+i)))=((1+i)/(1−i^2 ))=((1+i)/(1−(−1))) =((1+i)/2)=(1/2)+(i/2) ⇒ real part of z = (1/2)
$$ \\ $$$$\mathrm{Q37}. \\ $$$$\mathrm{Rationalising}\:{z} \\ $$$$\Rightarrow\:\frac{\mathrm{1}\left(\mathrm{1}+{i}\right)}{\left(\mathrm{1}−{i}\right)\left(\mathrm{1}+{i}\right)}=\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}^{\mathrm{2}} }=\frac{\mathrm{1}+{i}}{\mathrm{1}−\left(−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}+{i}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{real}\:\mathrm{part}\:\mathrm{of}\:\mathrm{z}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by AST last updated on 24/Dec/20
Q38. 4+a−i=2b+ai Comparing real and imaginary part ⇒ a=−1 and 4+a=2b ⇒ 4−1=2b ⇒ b = (3/2) Hence, (a,b) = (−1, (3/2))
$$\mathrm{Q38}. \\ $$$$\mathrm{4}+\mathrm{a}−{i}=\mathrm{2}{b}+{ai} \\ $$$${C}\mathrm{omparing}\:\mathrm{real}\:\mathrm{and}\:\mathrm{imaginary}\:\mathrm{part} \\ $$$$\Rightarrow\:\mathrm{a}=−\mathrm{1}\:\mathrm{and}\:\mathrm{4}+\mathrm{a}=\mathrm{2b} \\ $$$$\Rightarrow\:\mathrm{4}−\mathrm{1}=\mathrm{2b}\:\Rightarrow\:\mathrm{b}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{Hence},\:\left(\mathrm{a},\mathrm{b}\right)\:=\:\left(−\mathrm{1},\:\frac{\mathrm{3}}{\mathrm{2}}\right)\: \\ $$$$ \\ $$
Answered by AST last updated on 24/Dec/20
Q39. a+ib=(2−i)^2 =4+i^2 −4i=4+(−1)−4i =3−4i Comparing real and imaginary parts a=3, b=−4 ⇒ a+b = 3+(−4)=−1
$$\mathrm{Q39}. \\ $$$$\mathrm{a}+{i}\mathrm{b}=\left(\mathrm{2}−{i}\right)^{\mathrm{2}} =\mathrm{4}+{i}^{\mathrm{2}} −\mathrm{4}{i}=\mathrm{4}+\left(−\mathrm{1}\right)−\mathrm{4}{i} \\ $$$$=\mathrm{3}−\mathrm{4}{i} \\ $$$$\mathrm{Comparing}\:\mathrm{real}\:\mathrm{and}\:\mathrm{imaginary}\:\mathrm{parts} \\ $$$$\mathrm{a}=\mathrm{3},\:\mathrm{b}=−\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{a}+\mathrm{b}\:=\:\mathrm{3}+\left(−\mathrm{4}\right)=−\mathrm{1} \\ $$
Answered by AST last updated on 24/Dec/20
Q40. Rationalising z ⇒ ((1(i))/(i(i)))=(i/i^2 )=(i/(−1))=−i Hence,z^− =i
$$\mathrm{Q40}.\: \\ $$$$\mathrm{Rationalising}\:\mathrm{z} \\ $$$$\Rightarrow\:\frac{\mathrm{1}\left({i}\right)}{{i}\left({i}\right)}=\frac{{i}}{{i}^{\mathrm{2}} }=\frac{{i}}{−\mathrm{1}}=−{i} \\ $$$$\mathrm{Hence},\overset{−} {\mathrm{z}}={i} \\ $$

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