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COMPLEX NUMBERS Class XI
Quadratic Equation:-
Q.1) Find all non-zero complex numbers π§ satisfying |π§| = ππ§2
Sol.1) Let π§ = π₯ + ππ¦ We have |π§| = ππ§2 β π₯ β ππ¦ = π(π₯ + ππ¦)2 β π₯ β ππ¦ = π(π₯2 β π¦2 + 2ππ₯π¦) β π₯ β ππ¦ = ππ₯2 β ππ¦2 β 2π₯π¦ β π₯ β ππ¦ = β2π₯π¦ + π(π₯2 β π¦2) Equating real & imaginary parts
β π₯ = β2π₯π¦ and β π¦ = π₯2 β π¦2 β π₯ + 2π₯π¦ = 0 and π₯2 β π¦2 + π¦ = 0 Consider π₯ + 2π₯π¦ = 0 β π₯(1 + 2π¦) = 0
β π₯ = 0 (or) π¦ =β1
2
Case 1 , when π₯ = 0 β π₯2 β π¦2 + π¦ = 0 β π¦(βπ¦ + 1) = 0 β π¦ = 0 (or) π¦ = 1 β΄ π₯ = 0 & π¦ = 0 and π₯ = 0 & π¦ = 1 β π§ = 0 + 0π and π§ = 0 + 1π (rejected) β΄ π§ = π
Case 2, when π¦ =β1
2
β π₯2 β π¦2 + π¦ = 0
β π₯2 β1
4β
1
2= 0
β π₯2 =3
4
β π₯ = Β±β3
2
β΄ π₯ =β3
2, π¦ =
β1
2 and π₯ =
ββ3
2, π¦ =
β1
2
β π§ =β3
2β
1
2π and π§ =
ββ3
2β
1
2π
β΄ required complex numbers are = π,β3
2β
1
2π and
ββ3
2β
1
2π ans.
Q.2) Solve the equation π§2 + |π§| = 0.
Sol.2) Let π§ = π₯ + ππ¦ We have π§2 + |π§| = 0
β (π₯ + ππ¦)2 + βπ₯2 + π¦2 = 0
β π₯2 β π¦2 + 2ππ₯π¦ + βπ₯2 + π¦2 = 0
β (π₯2 β π¦2 + βπ₯2 + π¦2) + π2π₯π¦ = 0 + 0π
β π₯2 β π¦2 + βπ₯2 + π¦2 = 0 and 2π₯π¦ = 0 Consider 2π₯π¦ = 0
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π₯ = 0 or π¦ = 0 Case 1, when π₯ = 0
Then π₯2 β π¦2 + βπ₯2 + π¦2 = 0
β βπ¦2 + βπ¦2 = 0
β βπ¦2 + |π¦| = 0 If π¦ > 0 then |π¦| = π¦ β βπ¦2 + π¦ = 0 β π¦(βπ¦ + 1) = 0 π¦ = 0 or π¦ = 1 If π¦ < 0 then |π¦| = βπ¦ β βπ¦2 + π¦ = 0 β βπ¦(π¦ + 1) = 0 π¦ = 0 or π¦ = β1 π₯ = 0; π¦ = 0 or π₯ = 0; π¦ = β1 β¦β¦β¦β¦ (ii) Case 2, when π¦ = 0
Then π₯2 β π¦2 + βπ₯2 β π¦2 = 0
β π₯2 + βπ₯2 = 0 β π₯2 + |π₯| = 0 If π₯ > 0 then |π₯| = π₯ β π₯2 + π₯ = 0 β π₯(π₯ + 1) = 0 π₯ = 0 or π₯ = β1 (rejected since π₯ > 0) π₯ = 0; π¦ = 0 β¦β¦β¦β¦β¦ (iii) If π₯ < 0 then |π₯| = βπ₯ β π₯2 β π₯ = 0 β π₯(π₯ β 1) = 0 π₯ = 0 or π₯ = 1 (rejected since π₯ < 0) π₯ = 0; π¦ = 0 β¦β¦β¦β¦β¦ (iv) From (i), (ii), (iii) & (iv) π§ = 0 + ππ; π§ = 0 + π; π§ = 0 β π ans.
Q.3) If |π§1| = |π§2| = |π§3| β¦ β¦ β¦ β¦ |π§π| = 1 show that |π§1 + π§2 + π§3 + β― β¦ . . π§π| = |1
π§1+
1
π§2+
1
π§3+
β― β¦ . .1
π§π|
Sol.3) We have, |π§1 + π§2 + π§3 + β― β¦ . . π§π|
= |π§1π§1
π§1 +
π§2π§2
π§2 +
π§3π§3
π§3 + β― β¦ . .
π§ππ§π
π§π| (multiply & divide)
=|π§1|2
π§1 +
|π§2|2
π§2 +
|π§3|2
π§3 + β― β¦ . .
|π§π|2
π§π β¦β¦β¦.. {π§π§ = |π§|2}
= |1
π§1 +
1
π§2 +
1
π§3 + β― β¦ . .
1
π§π| β¦β¦β¦β¦.. {|π§1| = |π§2| = 1}
= |1
π§1+
1
π§2+
1
π§3+ β― β¦ . .
1
π§π
| β¦β¦β¦β¦β¦.. {π§1 + π§2 = π§1 + π§2}
= |1
π§1+
1
π§2+
1
π§3+ β― β¦ . .
1
π§π| β¦β¦.. |π§| = |π§|
Q.4) If |π§2 β 1| = |π§|2 + 1, show that π§ lies on imaginary axis.
Sol.4) Let π§ = π₯ + ππ¦
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Then |π§2 β 1| = |π§|2 + 1
= |(π₯ + ππ¦)2 β 1| = βπ₯2 + π¦22
+ 1 = |(π₯2 β π¦2 β 1) + 2ππ₯π¦| = (π₯2 + π¦2) + 1
= β(π₯2 β π¦2 β 1) + 4π₯2π¦2 = π₯2 + π¦2 + 1 Squaring both sides = (π₯2 β π¦2 β 1)2 + 4π₯2π¦2 = (π₯2 + π¦2 + 1)2 = π₯4 + π¦4 + 1 β 2π₯2π¦2 + 2π¦2 β 2π₯2 + 4π₯2π¦2
= π₯4 + π¦4 + 1 + 2π₯2π¦2 + 2π¦2 + 2π₯2 + 4π₯2π¦2 = 4π₯2 = 0 = π₯2 = 0 = π₯ = 0 π§ = 0 + ππ¦ Clearly π§ lies on π¦ β ππ₯ππ
Q.5) If the imaginary part of 2π§+1
ππ§+1 is β2, then show that the locus of the point representing π§ in
the argand plane is a straight line. LOCUS is a path traced by a moving point (π₯, π¦)
Sol.5) Let π§ = π₯ + ππ¦ Here, we have prove that equation containing π₯ & π¦ must be linear i.e., straight line
Now 2π§+1
ππ§+1=
2(π₯+ππ¦)+1
π(π₯+ππ¦)+1
=(2π₯+1)+π2π¦
(1βπ¦)+ππ₯
Rationalize
=(2π₯ + 1) + π2π¦
(1 β π¦) + ππ₯Γ
(1 β π¦) β ππ₯
(1 β π¦) β ππ₯
=(2π₯ + 1)(1 β π¦) β ππ₯(2π₯ + 1) + π2π¦(1 β π¦) β π22π₯π¦
(1 β π¦)2 β π2π₯2
=2π₯ β 2π₯π¦ + 1 β π¦ β π(2π₯2 + π₯) + π(2π¦ β 2π¦2) + 2π₯π¦
1 + π¦2 β 2π¦ + π₯2
=(2π₯ + 1 β π¦)
π₯2 + π¦2 β 2π¦ + 1+ π
(2π¦ β 2π¦2 β 2π₯2 β π₯)
π₯2 + π¦2 β 2π¦ + 1
Here πΌπ (2π§+1
ππ§+1) =
2π¦β2π¦2β2π₯2βπ₯
π₯2+π¦2β2π¦+1
Given πΌπ (2π§+1
ππ§+1) = β2
= 2π¦β2π¦2β2π₯2βπ₯
π₯2+π¦2β2π¦+1= β2
β 2π¦ β 2π¦2 β 2π₯2 β π₯ = β2π₯2 β 2π¦2 + 4π¦ β 2 β π₯ + 2π¦ β 2 = 0 Clearly this represents the equation of straight line
Q.6) Let π§1 & π§2 be two complex numbers such that |π§1 + π§2| = |π§1| + |π§2| then show that arg(π§1) βarg(π§2) = 0.
Sol.6) Let π§1 = π1 (cos π1 + π sin π1) And π§2 = π2 (cos π2 + π sin π2) Here |π§1| = π1 arg(π§1) = π1 β |π§2| = π2 arg(π§2) = π2
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We have, |π§1 + π§2| = |π§1| + |π§2| β |π1(cos π1 + π sin π1) + π2(cos π2 + π sin π2)| = π1 + π2 β |(π1 cos π1 + π2 cos π2) + π(π1 sin π1 + π2 sin π2)| = π1 + π2
β β(π1 cos π1 + π2 cos π2)2 + (π1 sin π1 + π2 sin π2)2 = π1 + π2 Squaring both sides β π1
2 cos2 π1 + π22 cos2 π2 + 2 π1π2 cos π1 cos π2 + π2
2 sin2 π2 + 2π1π2 sin π1 sin π2 =(π1 + π2)2 β π1
2(cos2 π1 + sin2 π1) + π22(cos2 π2 + sin2 π2) + 2π1π2(cos π1 cos π2 + sin π1 sin π2) =
π12 + π2
2 + 2π1π2 β π1
2 + π22 + 2π1π2 cos(π1 β π2) = π1
2 + π22 + 2π1π2
β cos(π1 β π2) = 1 β π1 β π2 = 0 β¦β¦β¦.. {cos π = 1 π‘βππ π = 0} βarg(π§1) βarg(π§2) = 0
Q.7) What does this equation |π§ + 1 β π| = |π§ β 1 + π| represents.
Sol.7) Let π§ = π₯ + ππ¦ Then |π§ + 1 β π| = |π§ β 1 + π| β |π₯ + ππ¦ + 1 β π| = |π₯ + ππ¦ β 1 + π| β |(π₯ + 1)π(π¦ β 1)| = |(π₯ + 1)π(π¦ + 1)|
β β(π₯ + 1)2 + (π¦ β 1)2 = β(π₯ β 1)2 + (π¦ + 1)2
β βπ₯2 + 1 + 2π₯ + π¦2 + 1 β 2π¦ = βπ₯2 + 1 β 2π₯ + π¦2 + 1 β 2π¦ Squaring β π₯2 + π¦2 + 2π₯ β 2π¦ + 2 = π₯2 + π¦2 + 2π¦ β 2π₯ + 2 β 4π₯ β 4π¦ = 0 β π₯ β π¦ = 0 Clearly this represents the equation of a straight line.
Q.8) Evaluate: 1 + π2 + π4 + π6 + β― β¦ β¦ . π20
SOL.8) = 1 β 1 + 1 β 1 + β― . +1 = 1 ans.
Q.9) Find the value of π4π+1βπ4πβ1
2
Sol.9) β π4π.πβπ4π.πβ1
2
β 1(π)β1(
1
1)
2 β¦β¦β¦β¦ {π4π = (π4)π = (1)π = 1}
β πβ(π)
2 β¦β¦β¦β¦.
1
1=
1
1Γ
π
π= βπ
β 2π
2= π
Q.10) What is the smallest positive integer π for which (1 + π)2π = (1 β π)2π.
Sol.10) We have, (1 + π)2π = (1 β π)2π
β (1+π)2π
(1βπ)2π = 1
β (1+π
1βπ)
2π= 1
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