TOPICS

Miscellaneous

Complex Numbers And Quadratic Equations

**Example-1 :-**

**Example-2 :-** Find the modulus and argument of the complex numbers:

(i) We have, (1+i)/(1-i) = (1+i)/(1-i) x (1+i)/(1+i) = (1+i²+2i)/(1-i²) = (1-1+2i)/(1+1) = 2i/2 = i = 0 + i Hence, z = 0 + i Now, 0 = r cos θ, 1 = r sin θ By squaring and adding, we get r^{2}cos^{2}θ + r^{2}sin^{2}θ = 0^{2}+ 1^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 0 + 1 r^{2}(cos^{2}θ + sin^{2}θ) = 1 r x 1 = √1 r = 1 Modulus = 1 Therefore, 0 = r cos θ and 1 = r sin θ cos θ = 0 and sin θ = 1, which gives θ = π/2 Argument = π/2

(ii) We have, 1/(1+i) = 1/(1+i) x (1-i)/(1-i) = (1-i)/(1-i²) = (1-i)/(1+1) = (1-i)/2 = 1/2 - i/2 We have, z = 1/2 - i/2 Now, 1/2 = r cos θ, -1/2 = r sin θ By squaring and adding, we get r^{2}cos^{2}θ + r^{2}sin^{2}θ = (1/2)^{2}+ (-1/2)^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 1/4 + 1/4 r^{2}(cos^{2}θ + sin^{2}θ) = 1/2 r x 1 = 1/√2 r = 1/√2 Modulus = 1/√2 Therefore, 1/2 = r cos θ and -1/2 = r sin θ cos θ = 1/√2 and sin θ = -1/√2, which gives θ = -π/4 Argument = -π/4

**Example-3 :-**

**Example-4 :-**

**Example-5 :-** Convert the complex number

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