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The Value Of 1 I 5 1 I 5

Solution: (1+i)5(1-i)5 = (1+i)2(1+i)2(1+i)(1-i)2(1-i)2(1-i)..(i) (1+i)2= 1+2i-1 = 2i (1-i)2 = 1-2i-1 = -2i (1+i)(1-i) = 1+1 = 2 Substitute... View Article

The Real Part Of 1 I I Is

(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

Real Part Of E Ei Theta Is

(1) ecos θ( cos (sin θ) (2) ecos θ( cos (cos θ) (3) esin θ( sin (cos θ) (4) esin θ( sin (sin θ) Solution: ecos θ (ei sin θ) = ecos θ(... View Article