Square Roots of Complex Numbers:
The square roots of complex numbers can be determined either by expressing the complex number in the form of a perfect square or by expressing the square roots in the form of a + ib and then evaluating the real numbers ‘a’ and ‘b’ by squaring both the sides and then equating the real and the imaginary parts.
| Let us assume, √(x + iy) = a + ib Then, x + iy = (a + ib)2 = a2 – b2 + 2abi ∴ Equating real and imaginary parts, we get- a2 – b2 = x ………(i) and 2ab = y ……..(ii) Now, (a2 + b2)2 = (a2 – b2)2 + 4a2b2 ⇒ (a2 + b2)2 = x2 + (2ab)2 = x2 + y2 ∴ a2 + b2 = +√(x2 + y2) [the negative sign is disregarded as x, y ∈ R] Thus, a2 – b2 = x and a2 + b2 = √(x2 + y2) |
It is clear from (ii) that both a and b must be of the same sign (either both positive or both negative) when y > 0 and a and b will have opposite signs if y < 0.
Thus, if y > 0, then the square roots of x + iy are-
Again, if y < 0, then the square roots of x + iy are-
| Example- If a2 + b2 = 1, show that (1 + ai + b)/(1 – ai + b) = b + ai Solution- The given equation is, (1 + ai + b)/(1 – ai + b) = b + ai Taking L.H.S., = (1 + ai + b)/(1 – ai + b) = [(1 + b) + ai]/[(1 + b) – ai] = [(1 + b) + ai]/[(1 + b) – ai] x [(1 + b) + ai]/[(1 + b) + ai] =[(1 + b) + ai]2/[(1 + b)2 – (ai)2] = [(1 + b)2 + (ai)2 + 2 (1 + b) (ai)]/[1 + b2 + 2b + a2] = [1 + b2 + 2b – a2 + 2 (1 + b) (ai)]/[1 + a2 + b2 + 2b] We know that a2 + b2 = 1, therefore equation becomes- = [a2 + b2 + b2 + 2b – a2 + 2 (1 + b) (ai)]/[1 + 1 + 2b] = [2b2 + 2b + 2ai (1 + b)]/[2 + 2b] = [2b (b + 1) + 2ai (1 + b)]/[2 (1 + b)] = [2 (1 + b) (b + ai)]/[2 (1 + b)] = b + ai Hence L.H.S. = R.H.S. |
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