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Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
Complex numbers
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Complex numbers

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  1. ASSESSMENT OF COMPLEX NUMBERS & QUADRATIC EQUATIONS (XI)<br /> Level...1<br /> Q.1 Solve X2 – (3√2 – 2i) x - 6√2i = 0 (3√2, -2i)<br /> Q.2 1+ix-2i3+i + 2-3iy+i3-i = i, find real values of x and y. <br /> Q.3 If z = (1+i1-i ), then z4 is<br /> (i) 1 (ii) -1 (iii) 0 (iv) none of them.<br /> Q.4 If z = 11-I(2+3I) , then |z| is<br /> (i) 1 (ii) 1/√(26) (iii) 5/√(26) (iv) none <br /> Q.5 If x+iy = 3+5i7-6i , then y = <br /> (a) 9/85 (b) -9/85 (c) 53/85 (d) -53/85<br /> Q.6 If z = 11-cosφ-isinφ , then Re(z) is <br /> (a) 0 (b) ½ (c) cot(φ/2) (d) (½)cot(φ/2)<br /> Level......2<br /> Q.1 Find the real values of ϴ for which the complex number 1+icosϴ1-2icosϴ is purely real.<br /> Q.2 If (1 - i) (1 - 2i)(1 - 3i)...........(1 - ni) = (x - yi) , show that 2.5.10...........(1+n2) = x2+y2 .<br /> Q.3 Prove that arg(z) = 2π – arg(z) ,z ≠0<br /> Q.4 Express in polar form: -21+i√3 . <br /> Q.5 If iz3 +z2 – z+ i = 0, then show that |z| = 1. <br /> Q.6 If a+ib = c+ic-i , then a2+b2 = 1 and b/a = 2cc²-1 <br /> Q.7 If x = - 5 +2√(-4) , find the value of x4+9x3+35x2 – x+4. <br /> Q.8 Show that a real x will satisfy equation 1-ix1+ix = a – ib, if a2+b2 = 1 where a, b are real.<br /> Q.9 A variable complex z is such that arg ( z-1z+1 ) = π2 , show that x2+y2 – 1=0<br /> Q.10 Find the values of x and y if x2 – 7x +9yi and y2i+20i – 12 are equal. <br /> Answers of Level—2<br /> 1. Rationalise it with 1+2icosϴ , then equate imaginary part to 0, value will be 2nπ±π2 , n ЄZ.<br /> 2. By taking modulus or conjugate on the both sides. <br /> 3. Let z = r( cosϴ + i sinϴ) , z = r (cosϴ - i sinϴ) = r (cos(2π-ϴ) + i sin(2π-ϴ).<br /> 4. (cos2π/3 +isin2 π/3)<br /> 5. take i outside, make factors as z2 (z – i)+ i (z – i) = 0 ⇨ z = i, z2= -i <br />|z| = |i| =1, |z2| =|-i|=1 ⇨ |z|2 = |z2| =1⇨ |z| = 1.<br />6. By taking conjugate on the both sides and use the given value.<br />7. Divide given poly. By x2 +10x +41=0 as (x+5)2= (4i)2 <br /> x2 +10x +41 x4+9x3 +35x2 – x+4<br /> x4+10x3 +41x2<br /> -x3 - 6x2 – x<br /> -x3 – 10x2 – 41x <br /> 4x2 + 40x + 4<br /> 4x2 + 40x +164<br /> -160 -> answer. <br /> <br /> 8. By C &D then we will get x = 2b1+a2²+b² .<br /> 9. Assume z = x+iy, arg ( z-1z+1 ) = π2 ⇨ arg( x+iy-1x+iy+1 ) = π2 ⇨ arg( x-1+iyx+1+iy x x+1-iyx+1-iy ) = π2 <br /> ⇨ tan-1π2 = 2xyx²+y²-1 .<br /> 10. x =4, 3 and y =5, 4.<br />it can be used in finding principal argument of complex numbers.<br /> ϴ, x>0, y>0<br /> π - ϴ, x<0, y>0 <br /> arg (z) = ϴ - π , x<0, y<0 <br /><ul><li> - ϴ, x>0, y<0

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