Solution: (1-i)n = 2n+0i If two complex numbers are equal, their moduli are also equal. |(1-i)n| = |2n| |1-i|n = |2n| √(12+(-1)2]n = 2n... View Article
(1) e-π/4 cos (½ log 2) (2) -e-π/4 sin (½ log 2) (3) eπ/4 cos (½ log 2) (3) e-π/4 sin (½ log 2) Solution: Let z = (1-i)-i On taking log... View Article
Solution: Given z = (1+i√3)/(√3+i) Multiply numerator and denominator with (√3-i) z = (1+i√3)(√3-i)/(√3+i)(√3-i) = (√3+3i-i+√3)/(3+1) =... View Article
Solution: (1+i√3)9 = a+ib = 29 [(½)+(√3/2)i ]9 = 29 [ cos π/3 + i sin π/3]9 Apply de moivre's theorem = 29 [ cos 9π/3 + i sin 9π/3] =... View Article
Solution: Let x3 = -i x3+i = 0 (x3-(i)3 )= 0 (x-i)(x2-1+ix) = 0 x - i = 0 or x2-1+ix = 0 x = i or x = -i±√(-1+4)/2 x = i or x = (-i±√3)/2... View Article
Solution: Given y = cos θ+i sin θ 1/y = 1/(cos θ+i sin θ) Multiply numerator and denominator with (cos θ-i sin θ) 1/y = (cos θ-i sin θ)/... View Article
Solution: Given -1+√-3 = reiθ -1+√-3 = -1+√3i -1+√3i = r(cos θ+i sin θ) r cos θ = -1 r sin θ = √3 Squaring and adding r2(cos2θ+sin2θ)... View Article
(1) er sin θ (2) e-r sin θ (3) e-r cos θ (4) er cos θ Solution: Given z = reiθ eiθ = cos θ+i sin θ So z = r(cos θ+i sin θ) iz... View Article
(1) √2 [cos (3π/4)+i sin (3π/4)] (2) √2 [cos (π/4)+i sin (π/4)] (3) cos (3π/4)+i sin (3π/4) (4) None of these Solution: (1+7i)/(2-i)2 =... View Article
Solution: Given eiθ = cos θ+ i sin θ In triangle A+B+C = 1800 = π eiA.eiB.eiC = ei(A+B+C) = eiπ = cos π+ i sin π = -1 Hence option... View Article
