## 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 + *i*b 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 + ibThen, x + iy = (a + ib)^{2} = a^{2} – b^{2} + 2abi∴ Equating real and imaginary parts, we get- a ^{2} – b^{2} = x ………(i) and 2ab = y ……..(ii)Now, (a ^{2} + b^{2})^{2} = (a^{2} – b^{2})^{2} + 4a^{2}b^{2}⇒ (a ^{2} + b^{2})^{2} = x^{2} + (2ab)^{2} = x^{2} + y^{2} ∴ a ^{2} + b^{2} = +√(x^{2} + y^{2}) [the negative sign is disregarded as x, y ∈ R]Thus, a ^{2} – b^{2} = x and a^{2} + b^{2} = √(x^{2} + y^{2}) |

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 + *i*y are-

Again, if y < 0, then the square roots of x + *i*y are-

Example- If a^{2} + b^{2} = 1, show that (1 + ai + b)/(1 – ai + b) = b + aiSolution- The given equation is, (1 + a i + b)/(1 – ai + b) = b + aiTaking L.H.S., = (1 + a i + b)/(1 – ai + b)= [(1 + b) + a i]/[(1 + b) – ai]= [(1 + b) + a i]/[(1 + b) – ai] x [(1 + b) + ai]/[(1 + b) + ai]=[(1 + b) + a i]^{2}/[(1 + b)^{2} – (ai)^{2}]= [(1 + b) ^{2} + (ai)^{2} + 2 (1 + b) (ai)]/[1 + b^{2} + 2b + a^{2}]= [1 + b ^{2} + 2b – a^{2} + 2 (1 + b) (ai)]/[1 + a^{2} + b^{2} + 2b]We know that a ^{2} + b^{2} = 1, therefore equation becomes-= [a ^{2} + b^{2} + b^{2} + 2b – a^{2} + 2 (1 + b) (ai)]/[1 + 1 + 2b] = [2b ^{2} + 2b + 2ai (1 + b)]/[2 + 2b]= [2b (b + 1) + 2a i (1 + b)]/[2 (1 + b)]= [2 (1 + b) (b + a i)]/[2 (1 + b)]= b + a iHence L.H.S. = R.H.S. |