ASSESSMENT OF COMPLEX NUMBERS & QUADRATIC EQUATIONS (XI)

Level...1 Q.1 Solve X2 – (3√2 – 2i) x - 6√2i = 0 (3√2, -2i)

Q.2 (1+ i ) x−2 i3+i + (2−3 i ) y+i

3−i = i, find real values of x and y.

Q.3 If z = ( 1+i1−i ), then z4 is

(i) 1 (ii) -1 (iii) 0 (iv) none of them.

Q.4 If z = 1

(1−I )(2+3 I ) , then |z| is

(i) 1 (ii) 1/√(26) (iii) 5/√(26) (iv) none

Q.5 If x+iy = 3+5 i7−6 i , then y =

(a) 9/85 (b) -9/85 (c) 53/85 (d) -53/85

Q.6 If z = 11−cosφ−isinφ , then Re(z) is

(a) 0 (b) ½ (c) cot(φ/2) (d) (½)cot(φ/2)

Level......2 Q.1 Find the real values of ϴ for which the complex number 1+ icosϴ

1−2 icosϴ is purely real.

Q.2 If (1 - i) (1 - 2i)(1 - 3i)...........(1 - ni) = (x - yi) , show that 2.5.10...........(1+n2) = x2+y2 . Q.3 Prove that arg(z) = 2π – arg(z) ,z ≠0

Q.4 Express in polar form: −21+ i√ 3 .

Q.5 If iz3 +z2 – z+ i = 0, then show that |z| = 1.

Q.6 If a+ib = c+ic−i , then a2+b2 = 1 and b/a = 2cc ²−1

Q.7 If x = - 5 +2√(-4) , find the value of x4+9x3+35x2 – x+4.

Q.8 Show that a real x will satisfy equation 1−ix1+ix =

a – ib, if a2+b2 = 1 where a, b are real.

Q.9 A variable complex z is such that arg ( z−1z+1 ) = π2

, show that x2+y2 – 1=0 Q.10 Find the values of x and y if x2 – 7x +9yi and y2i+20i – 12 are equal.

Answers of Level—2 1. Rationalise it with 1+2icosϴ , then equate imaginary part to 0, value will be 2nπ±

π2 , n Z.Є

2. By taking modulus or conjugate on the both sides. 3. Let z = r( cosϴ + i sinϴ) , z = r (cosϴ - i sinϴ) = r (cos¿

+ i sin ¿. 4. (cos2π/3 +isin2 π/3) 5. take i outside, make factors as z2 (z – i)+ i (z – i) = 0 ⇨ z = i, z2= -i

|z| = |i| =1, |z2| =|-i|=1 ⇨ |z|2 = |z2| =1⇨ |z| = 1.6. By taking conjugate on the both sides and use

the given value.7. Divide given poly. By x2 +10x +41=0 as

(x+5)2= (4i)2

x2 +10x +41 x4+9x3 +35x2 – x+4 x4+10x3 +41x2

-x3 - 6x2 – x -x3 – 10x2 – 41x 4x2 + 40x + 4 4x2 + 40x +164 -160 → answer.

8. By C &D then we will get x = 2b

(1+a2 ) ²+b ² .

9. Assume z = x+iy, arg ( z−1z+1 ) = π2 ⇨ arg( x+iy−1

x+ iy+1 ) = π2 ⇨ arg( x−1+iy

x+1+iy xx+1−iyx+1−iy ) = π

2

⇨ tan−1 π2 =

2xyx ²+ y ²−1 .

10. x =4, 3 and y =5, 4.it can be used in finding principal argument of complex numbers. ϴ, x>0, y>0 π - ϴ, x<0, y>0 arg (z) = ϴ - π , x<0, y<0 - ϴ, x>0, y<0