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Question-126827




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)
Q33.4+7i13iRationalising(4+7i)(1+3i)(13i)(1+3i)=4+12i+7i+(21i2)12(3i)2=4+19i211(9)=17+19i10=1710+i(1910)=1.7+1.9i(C)
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)
Q34.Rationalising1+i2+i(1+i)(2i)(2+i)(2i)=2i+2ii24i2=2+i(1)4(1)=3+i5=35+15ix=35
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
Q35.Rationalisingz1(12i)(1+2i)(12i)=12i1(2i)2=12i1(4)=12i5=1525iz=15+25i
Answered by AST last updated on 24/Dec/20
Q36.  Note that z^�_−   = z  ⇒ conjugate of (conjugate of z) = z  Hence, answer is B.
Q36.Notethatz¯=zconjugateof(conjugateofz)=zHence,answerisB.
Commented by harckinwunmy last updated on 24/Dec/20
thanks AST. can you help complete it
thanksAST.canyouhelpcompleteit
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)
Q37.Rationalisingz1(1+i)(1i)(1+i)=1+i1i2=1+i1(1)=1+i2=12+i2realpartofz=12
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))
Q38.4+ai=2b+aiComparingrealandimaginaryparta=1and4+a=2b41=2bb=32Hence,(a,b)=(1,32)
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
Q39.a+ib=(2i)2=4+i24i=4+(1)4i=34iComparingrealandimaginarypartsa=3,b=4a+b=3+(4)=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
Q40.Rationalisingz1(i)i(i)=ii2=i1=iHence,z=i

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