PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1wqYpF34FSOPB8FKkG-3VTIuK2XunsImF/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true Complex Numbers Complex numbers as ordered pairs of reals Representation of complex numbers in the form a+ib and their representation in a plane, Argand diagram, algebra of complex numbers, modulus and argument (or amplitude) of a complex number, square root of a complex C H A P T E R number, triangle inequality. CHAPTER IMAGINARY NUMBERS Square root of a negative number is called imaginary number. While solving INCLUDES Algebra of Complex the equations x2 + 1 = 0, a quantity which is imaginary. is obtaind and is denoted by i (iota) numbers Addition Further is an imaginary number and can be written as Multiplication = 2 × = If a < 0, then = i i conjugation Properties of modulus and Integral Powers of i We have i = so i 2 = – 1, i 3 = –i, i 4 = 1 principal argument Polar-representation For any n ∈ N, we have i 4n + 1 = i , i 4n+2 = – 1 i 4n + 3 = –i, i 4n = 1 Thus any integral power of i can be expressed as ±1 or ±i. Triangle inequality Geometric interpretation Solved examples ⎪⎧(−1)n / 2 if n is even integer In other words i n = ⎨ ⎪⎩(–i ) n – 1 2 . i if n is odd integer Also i −n = 1 i n Evaluate: i 786 (− − 1)23 (iii) i 2 + i 3 + i 4 + i 5 i + i 2 + i 3 Solution : (i) i 786 = i 4×196+2 = i 2 = –1 (ii) (− − 1)23 = (− 1× i )23 = (− i )23 = − (i )23 = − (i )4×5+3 = −i 3 = −(−i ) = i (iii) i 2 + i 3 + i 4 + i 5 i + i 2 + i 3 = i 2(1+ i + i 2 + i 3 ) = i + i 2 + i 3 (− 1)(1+ i − 1− i ) i − 1− i = 0 = 0 − 1 COMPLEX NUMBERS The formal addition, ‘a + ib’, where a, b ∈ R and the collection of all such expressions is called the set of complex numbers. For the complex number, z = a + ib, ‘a’ is called as real part of z and is denoted by Re(z) while ‘b’ is called as imaginary part of z and is denoted by Im(z). Set of complex numbers is denoted by C, which includes the set of real numbers R. i.e., R ⊂ C A complex number z is said to be purely real if Im(z) = 0 and is said to be purely imaginary if Re(z) = 0. The complex number 0 + 0i = 0 is both purely real and purely imaginary. All purely imaginary numbers except zero are imaginary numbers but an imaginary number may or may not be purely imaginary. For e.g., 4 + 3i is imaginary but not purely imaginary. Equality of Complex numbers Two complex numbers a + ib and c + id are said to be equal, if and only if, a = c and b = d. i.e., the corresponding real and imaginary parts are equal. If a + ib = C1 and c + id = C2 then either C1 = C2 or C1 ≠ C For imaginary numbers, the property of order is not defined because i is neither positive, zero nor negative. So C1 > C2 or C2 > C1 is meaningless till b and d are both equal to zero. i.e., C1 > C2 or C1 < C2 are meaningless if b and d are not equal to zero. ALGEBRA OF COMPLEX NUMBERS Addition : (a + ib) + (c + id) = (a + c) + i (b + d) Subtraction : (a + ib) – (c + id) = (a – c) + i (b – d) Multiplication : (a + ib) · (c + id) = ac + iad + ibc + i2bd = (ac – bd) + i(ad + bc) Division : a + ib = ac + bd + i (bc − ad ) (When atleast one of c and d is non-zero) c + id Conjugate Complex Number c 2 + d 2 c 2 + d 2 For z = a + ib, its conjugate is defined as z = a − ib . Here the complex conjugate is obtained just by changing sign of i. Properties of conjugate : (z) = z z = z = iff z is purely real z + z = 2Re(z) ⇒ Re(z) = Re(z ) = z + z 2 z – z = 2i Im(z) ⇒ I (z) = z − z m 2i z = – z iff z is purely Imaginary (vi) (vii) z1 ± z2 = z1 ± z2 z1 z2 = z1 z2 ⎛ z1 ⎞ z1 z ⎜ z ⎟ = , 2 ≠ 0 ⎜ ⎟ ⎝ ⎠ (ix) ⎛⎜ zn ⎞⎟ ⎝ ⎠ = (z)n (x) z1 z2 + z1 z2 = 2Re(z1z2 ) = 2Re(z1z2 ) (xi) If z = f (z1), then z = f (z1) Modulus of a Complex Number The modulus of a complex number z = x + iy is defined as | z |= = {Re(z)}2 + {Im(z)}2 . In other way distance of a complex number z from origin while represented on argand plane is called as modulus of a complex number denoted by mod(z) or |z|, or r. Here OP = r = |z| is also called absolute value of z. Properties of Modulus : (i) |z| ≥ 0; |z| = 0 iff real and imaginary parts are zero. (ii) z1z2 = z1 z2 . In general z1z2......zn = z1 z2 zn (iii) = z1 , (z z2 ≠ 0) z = = − z = − z zz = | z |2 (vi) z ≤ Re(z) ≤ z , − z ≤ Im(z) ≤ z zn = z n (viii) z1 + z2 2 = (z z2 )(z1 + z2 ) = z1z1 + z1z2 + z2 z1 + z2z2 = | z1 |2 + | z2 |2 + 2Re (z1z2 ) (ix) z1 − z2 2 = (z z2 )(z1 − z2 ) = z1z1 − z1z2 − z2z1 + z2 z2 = |z |2 + |z |2 – 2Re (z1z2 ) (x) 1 2 z + z 2 + z − z 2 = 2 ⎛⎜ z 2 + z 2 ⎞⎟ 1 2 1 2 ⎝ 1 2 ⎠ (xi) |az – bz |2 + |bz + az |2 = (a2 + b2) (|z |2 + |z |2) where a, b ∈ R 1 2 1 2 1 2 Representation of a Complex Number Geometrical Representation The complex number, z = x + iy can be associated with the ordered pair P(x, y). We consider two perpendicular lines OX and OY (analogous to the Cartesian system) called as the real axis and imaginary axis respectively and where O denotes the origin of reference. The resulting plane is called as Argand plane or Gaussian plane or complex plane and z is represented by the point P corresponding to the ordered pair (x, y). The point P is called as affix of z. (Imaginary Axis) X (Real Axis) Argument or Amplitude of z The argument of z, denoted by arg z or amp z is the angle which OP makes with the positive direction of real axis, the angle being measured in anticlockwise sense. y If z = x + iy then the angle θ given by tanθ = x is said to be the argument of z. –1 ⎛ y ⎞ = tan−1⎛ Im(z) ⎞ or arg(z) = amp(z) = tan ⎜ x ⎟ ⎜ Re(z) ⎟ ⎝ ⎠ ⎝ ⎠ Principal argument of a complex number z is that value of argument (z) which lies in the interval (–π, π]. For z = x + iy, θ = tan−1 , then principal argument depends on the quadrant in which point (x, y) lies. x > 0, y = 0 ⇒ Principal argument = 0 (Positive Real Axis) x > 0, y > 0 ⇒ Prinicpal argument = θ (Ist quadrant) π x = 0, y > 0 ⇒ Prinicpal argument = 2 (Positive imaginary axis) x < 0, y > 0 ⇒ Principal argument = π – θ (IInd quadrant) x < 0, y = 0 ⇒ Prinicpal argument = π (Negative Real Axis) x < 0, y < 0 ⇒ Principal argument = –π + θ (IIIrd quadrant) x = 0, y < 0 ⇒ Principal argument = − π 2 (Negative Imaginary Axis) x > 0, y < 0 ⇒ Principal argument = –θ (ivth quadrant) Properties of Argument (i) arg (z1 z2 ) = arg (z1) + arg (z2 ) + 2k π : ⎛ z1 ⎞ k = 0, ± 1 (ii) arg ⎜ ⎝ z ⎟ = arg (z1) − arg (z2 ) + 2k π : 2 ⎠ k = 0, ± 1 (iii) arg (z2 ) = 2arg (z) + 2k π : k = 0, ± 1 (iv) If arg (z) = 0 or π then z is real (v) arg ⎛ z ⎞ = 2 arg(z) + 2kπ : k = 0, ± 1 ⎜ ⎟ (vi) ⎝ z ⎠ arg(zn ) = n arg(z) + 2kπ : k = 0, ± 1 (vii) ⎛ z2 ⎞ ⎛ z1 ⎞ If arg ⎜ z ⎟ = θ, then arg ⎜ z ⎟ = 2kπ − θ; k ∈ I ⎝ 1 ⎠ ⎝ 2 ⎠ arg (z ) = −arg(z) arg (z − z ) = ± π 2 arg(z) – arg (–z) = ±π Polar form (Trigonometric form) of a Complex Number Let OP = r, then x = r cosθ, and y = r sinθ ⇒ z = x + iy = r cosθ + ir sinθ, = r(cos θ + i sinθ). This is known as Trigonometric (or Polar) form of a complex Number. Here we should take the principal value of θ. For general values of the argument z = r[cos (2nπ + θ) + i sin(2nπ + θ)] (where n is an integer) Exponential form of a Complex Number (Euler’s Form) According to Euler’s Theorem, eiθ = cosθ + isinθ and therefore z = r(cosθ + isinθ) can be written as z = reiθ which is called as exponential form of a complex number. Replacing θ by –θ in eiθ, we obtain e–iθ = cosθ – isinθ Hence cos θ = eiθ + e−iθ 2 x x 2 + y 2 and sinθ = eiθ − e −iθ 2i y x 2 + y 2 Write the following complex numbers in polar and exponential form (1) − 1 − 3 i 2 2 (2) 1 – i Solution : Here rcosθ 1 θ = – 3 = – 2 , rsin 2 1 3 Squaring and adding r2cos2θ + r 2sin2θ = ∴ r = 1 Dividing we get tanθ = + = 1 4 4 ⎛ Since ⎜− 1 , − ⎞ ⎟ lies in third quadrant ⎜ ⎟ ⎝ ⎠ Principal argument = π − π = − 2π 3 3 1 3 ⎛ ⎛ 2π ⎞ ⎛ 2π ⎞⎞ 2π i ∴ Polar form of − − 2 i is 1 ⎜cos ⎜− 2 ⎟ + i sin⎜− 3 ⎟⎟ and Euler’s form is 1.e 3 3 ⎝ ⎝ ⎠ Here rcosθ = 1 and rsinθ = –1 ⎝ ⎠⎠ ∴ r = θ = –1 = tan⎛− π ⎞  , tan ⎜ ⎟ ⎝ 4 ⎠ Since (1, –1) lies in IVth quadrant, principal value of θ is − π 4 ∴ Polar form of 1 – i is ⎛ ⎛− π ⎞ +  ⎛− π ⎞⎞ 2 ⎜cos⎜ ⎟ 4 i sin⎜ ⎟⎟ 4 − iπ ⎝ ⎝ ⎠ ⎝ ⎠⎠ iπ Remember : 1 = ei0, i = e 2 , –i = e − iπ 2 , –1 = eiπ iπ π ⎛ π ⎞ π π π log i = log e 2 = i ; log(log i) = log⎜ i ⎟ = log i + log = i + log 2 ⎝ 2 ⎠ 2 2 2 Vector representation of a Complex Number A complex number z = x + iy can be represented by the position vector OP of point P(x, y) in a two dimensional plane becuase a complex number depends on two things viz (i) its modulus and (ii) its argument which are also the requirements of a vector on a plane. Geometrical meaning of Algebraic Operations y x′ O y′ P(x, y) x Let z1 = x1 + i y1 and z2 = x2 + i y 2 be represented by the points P1(x1, y1) and P2 (x2, y 2 ) respectively. y P(z + z ) 1 2 O R(–z2) P1(z1) x Q(z1–z2) Then, z1 + z2 = (x1 + x2 ) + i(y1 + y 2 ) is represented by the point P (where OP1 P P2 is a parallelogram). Similarly z1 − z2 = z1 + (−z2 ) is represented by the point Q (as shown). In vector notation z1 + z2 = OP1 + OP2 = OP1 + P1P = OP and z1 – z2 = OP1 − OP2 = OP1 + OR = OQ If z = x + iy, then, the conjugate of z, denoted by z is the number z = x − i y , x which is the mirror image of z along real axis. TRIANGLE INEQUALITY In any triangle, sum of any two sides is greater than the third side and difference of any two sides is less than the third side, we have ⎛ z1 ⎞ (i) z1 + z2 ≥ z1 + z2 ; here equality holds when arg⎜ ⎟ ⎝ z2 ⎠ = 0 i.e., position vectors representing two complex numbers z1 and z2 are parallel. (ii) z − z ≤ z − z ; here equality holds when ⎛ z1 ⎞ = 0 i.e., position vectors representing two complex 1 2 1 2 arg⎜ ⎟ ⎝ z2 ⎠ numbers z1 and z2 are parallel. Parallelogram law : In the parallelogram OP1PP2, the sum of the squares of its sides is half of the sum of the squares of its diagonals ⇒ z + z 2 + z − z 2 = 2⎛⎜ z 2 + z 2 ⎞⎟ 1 2 1 2 ⎝ 1 2 ⎠ Square Root of a Complex Number: Let z = x + iy be a complex number such that x + iy = a + ib where x, y ∈ R = a + ib ⇒ (a + ib)2 = x + iy ⇒ a2 – b2 + 2abi = x + iy On equating real and imaginary parts, x = a2 – b2 (i) and y = 2ab (ii) Solving (i) and (ii), we get a = ± ) and b = ± 1 ( 2 x ) ⎧ ⎡ | z | + x ⎤ ⎪± ⎢ 2 ⎥ + i for y > 0 Hence x + iy = ⎪ ⎢⎣ ⎥⎦ | z | − x ⎤ − i ⎥ for y < 0 2 ⎥⎦ Remember : ⎛ 1 + i ⎞ ⎛ 1 − i ⎞ = ± ⎜ ⎟, ⎝ ⎠ = ± ⎜ ⎟ ⎝ ⎠ Application of Complex Numbers to Geometrical Application Distance Formula : Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex number represented by points P and Q on the argand plane. Then the distance PQ = Also z2 – z1 = (x2 + iy2) – (x1 + iy1) = (x2 – x2) + i(y2 – y1) ∴ | z2 − z1 | = = PQ Hence distance between two points z1 and z2 is given by |z2 – z1| If z1 and z2 are two fixed points in the argand plane then find the locus of a point z in each of the following (i) |z – z1| + |z – z2| = |z1 – z2| (ii) |z – z1| = |z – z2| (iii) |z – z1| – |z – z2| = |z1 – z2| Solution : Let P and Q be two points represented by z1 and z2 in the argand plane and let R be a point having affix z. Then PR = |z – z1|, QR = |z – z2| and PQ = |z1 – z2| |z – z1| + |z – z2| = |z1 – z2| ⇒ PR + QR = PQ ⇒ R(z) lies on the line segment joining P(z ) and Q(z ) 1 2 (ii) |z – z1| = |z – z2| ⇒ PR = QR ⇒ R(z) is equidistant from P(z ) and Q(z ). 1 2 ⇒ R lies on the perpendicular bisector of line segment PQ. (iii) |z – z1| – |z – z2| = |z1 – z2| ⇒ PR – QR = PQ ⇒ R(z) lies on the line joining P(z ) and Q(z ) but does not lie between them. 1 1 Section formula : If the point R(z) divides the line segment PQ (where P and Q represent the complex numbers z1 and z2 respectively) mz + nz m n Internally in the ratio m : n then, Externally in the ratio m : n then, z = 2 1 m + n z = mz2 − nz1 P(z1) R(z) Q(z2) Mid point of line segment PQ = z1 + z2 2 m − n P(z1) Q(z2) R(z) Area of Triangle If the vertices of a triangle ABC represent the complex numbers z1, z2 and z3 respectively then area of the triangle is the modulus of z1 z1 1 z2 z2 1 . z3 z3 1 Show that the area of the triangle on the Argand diagram formed by the complex numbers z, iz and z + iz is z 2 . Solution : Let z = x + iy then co-ordinates in ordered pair of z and iz are (x, y) and (–y, x). Here origin 0, z, z + iz and iz form a square of side z . C(z + iz) Hence area of the required triangle is 1 z 2 . (Half the area of the square) 2 Slope of line segment joining two points If P and Q represent complex numbers z1 and z2 respectively in the Argand plane, then the complex slope of PQ is defined to be z1 − z2 . z1 − z2 Condition for collinearity: Three points A(z1) B(z2) and C(z3) will be collinear If there exists a relation az1 + bz2 + cz3 = 0, such that a + b + c = 0 (where a, b, c are non-zero real numbers) (where a, b, c are not all zero real numbers) If the area of ΔABC formed by z , z and z is zero z1 i.e., z2 z3 z1 1 z2 1 = 0 z3 1 1 2 3 If slope of AB = slope of BC = slope of AC i.e., z1 − z2 = z2 − z3 = z1 − z3 z1 − z2 z2 − z3 z1 − z3 Equation of a Straight Line Let P(z1) and Q(z2) be two given points in the Argand plane. If R(z) be any point on line joining P and Q, then we have arg (z – z2) = arg (z2 – z1) ⎛ z − z2 ⎞ i.e., arg⎜ z ⎟ = 0 z ⎝ 2 1 ⎠ z − z2 ⎛ z − z2 ⎞ ⇒ z − z − ⎜ z − z ⎟ = 0 (If arg (z) = 0 then z is real i.e., z – z = 0) 2 1 ⎝ 2 1 ⎠ ⇒ z − z2 − z − z2 = 0 z2 − z1 z2 − z1 or z(z1 − z2 ) − z(z1 − z2 ) + (z1 z2 − z2 z1) = 0 ....(i) z which can also be written as z1 z2 z 1 z1 1 = 0 z2 1 ....(ii) If in equation (i), z1 − z2 = a then z1 – z2 = a ⇒ Equation of straight line is of the form za − za + b = 0 (iii) where b = z1z2 − z2 z1 , which is purely imaginary. However, if equation (iii) is multiplied by i, it reduces to z(ia ) − z(ia) + (ib) = 0 or z(A ) + zA + B = 0 .....(iv) (Θ – ia, i a are conjugates) where B = ib is purely real Hence any equation of form (i), (ii), (iii) or (iv) will represent a straight line. Examples : 3z + 3z + 2 = 0 is a straight line since it is of form (iv) 3z – 3z – 6i = 0 is also a stright line of form (iii). 3z + 3z + 6i = 0 is not a straight line. For the straight line az + az + b = 0, b ∈ R Real slope of the line = a + a i (a − a ) = − Re(a) Im(a) and complex slope = − a a Distance of a point from a given line Distance of a point P(z ) from line za + za + b = 0 is given by | z1a + z1a + b | = | z1a + z1a + b | 2 | a | Condition for perpendicular or parallel lines Let complex slope of two lines be α and α For perpendicular lines, α + α = 0 For parallel lines, α = α ∴ Lines az + az + k1 = 0 and bz + bz + k2 = 0 (k , k ∈ R) are perpendicular if − a ⎛ + ⎜− ⎞ ⎟ = 0 1 2 or ab + ab = 0 a ⎝ ⎠ Concept of rotation Multiplication of a complex number z represented by point P, with eiθ = (cosθ + i sinθ) rotates the line OP by an angle θ, anticlockwise about O. Let z , z , z be the affixes of the vertices of a ΔABC described in the counter-clockwise sense. Draw OP 1 2 3 and OQ parallel and equal to AB and AC respectively. Then the point P is z2 – z1 and Q is z3 – z1 and z3 − z1 = OQ (cos α + i sinα) z2 − z1 OP = CA e iα = BA z3 − z1 eiα z2 − z1 ⎛ z3 − z1 ⎞ or amp⎜ z ⎟ = α z ⎝ 2 1 ⎠ C(z3) B(z2) In this case, we are rotating OP in clockwise direction by an angle (2π – α). Since the rotation is in clockwise direction, we are taking negative sign with angle (2π – α). Equation of the Perpendicular Bisector The equation of perpendicular bisector of the line segment joining points A(z1) and B(z2) is z − z1 = z − z2 or z(z1 z2 ) + z(z1 z2 ) = z1 2 − z 2 Properties of a triangle Let A(z1), B(z2), C(z3) are the vertices of a triangle, then : Centroid z1 + z2 + z3 3 If ABC is an equilateral triangle, then the circumcentre (z0) satisfies the relation z2 + z2 + z2 = 3z2 1 2 3 0 Incentre = If ABC is an equilateral triangle than z2 + z2 + z2 = z z + z z + z z 1 2 3 1 2 or 1 + 1 + 2 3 3 1 1 = 0 z1 − z2 z2 − z3 z3 − z1 If ABC is an isosceles right-angled triangle right angled at B then (z1 − z2 )2 = 2(z1 − z3 )(z3 − z2 ) and z2 + z2 + z2 = 2z (z z ) 1 2 3 Circle: 2 1 3 Circle with center represented by the complex number z0 and radius r has the equation | z–z0| = r. The general equation of a circle is zz + az + az + b = 0, where b is a real number. The center of this circle is ‘–a’ and its radius is . The equation of the circle described on the line segment joining z1 and z2 as diameter is (z − z1)(z − z2 ) + (z − z2 )(z − z1) = 0 . The point with affix z1 lies inside (resp. outside) the circle zz + az + az + b = 0 if z1z1 + az1 + az1 + b < 0 (resp. > 0). ⎛ z − z1 ⎞ If z is a variable point such that an arc of a circle. arg ⎜ ⎝ z − z ⎟ = α 2 ⎠ (where α ≠ nπ, α is a constant) then z describes (i) z − z1 z − z2 = k is a circle, if k ≠ 1 and is a line, if k = 1. The equation z − z1 2 + z − z 2 = k represents a circle if k ≥ z1 − z2 . If arg ⎡ (z2 − z3 )(z1 − z4 )⎤ = ±π, 0 , then the points z , z , z are concyclic. ⎢ (z − z )(z − z )⎥ 1 2 3 ⎣ 1 3 2 4 ⎦ (iv) z − z0 < r represents interior of the circle |z – z0| = r and |z – z0| > r represents exterior of the circle |z – z0| = r Useful Result If z + = a , the greatest and least valves of |z| are respectively and 2 2 SOLVED EXAMPLES Find the locus of complex number z if = 1. Solution : Given = 1 ⇒ = 1 ⇒ | i 3 | = 1 ⇒ |z + i | = |z – i | Which represents the equation of perpendicular bisector of i and –i and that will be the x-axis. Example 2 : If f(z) is divided by (z – i) and (z + i), the remainders are respectively i and 1 + i. Determine the remainder when f(z) is divided by z2 + 1. Solution : f(z) = (z– i)Q1(z) + i and f(z) = (z + i)Q2(z) + 1+ i where Q1(z) and Q2(z) are the quotients where f(z) is divided by (z – i) and (z + i) respectively. ∴ f(i) = i and f(–i) = 1+ i We have to determine the remainder when f(z) is divided by z2+ 1. Since the divisor is of degree 2, the remainder must be of degree less than 2 (i.e. either 1 or 0). Then f(z) = (z2 +1)Q(z) + pz + q : p, q ∈ C. ⇒ f(z) = (z – i) (z + i)Q(z) + pz + q pi + q = f (i) = i ... (1) –pi + q = f(–i) = 1+ i ... (2) Solving (1) and (2) for p and q , we have : p = i , 2 q = 1+ 2 i 2 Hence, the remainder is iz + 1+ 2i . 2 2 Find the least and greatest value of |z| which satisfies z + = a , where a ∈ R. Solution : Let z = r(cosθ + isinθ) then 1 = 1 (cos θ − i sin θ) z r a = z + = z − ⎛− 1 ⎞ ≥ z − − a ≤ r − 1 r ⎜ ⎟ ⎝ z ⎠ ≤ a r 2 + ar − 1 ≥ 0 ⎛ and r 2 − ar − 1 ≤ a − 0 a2 + 4 ⎞ ⎜ or r ≤ ⎜ 2 ⎝ ⎟ and 2 ⎟ ⎠ ≤ r ≤ 2 2 ∴ − a + rmax. a2 + 4 r ≤ 2 = ; 2 2 rmin. = 2 .