The usual quadratic formula follows from al Khwarizmi\'s identity: if p(z)= az^2

Question

The usual quadratic formula follows from al Khwarizmi's identity: if p(z)= az^2 + bz + c = a(z + b 2a)^2 - (b^2 - 4ac) 4a =0 then (z + b 2a)^2 = So if we can find a number r satisfying r^2=b^2-4ac, then we can solve for z in terms of a, b, c and r as z=-b 2a plusorminus r 2a. Hopefully you can recognize the usual quadratic formula here. If b^2 - 4ac is a positive real number, we usually just replace r with squareroot b^2-4ac. However if b^2 - 4ac is a negative real number, or in fact a general complex number---which will happen if a, b and c are complex numbers---then there is no canonical squareroot b^2 -4ac, but we could still find a value r (at least approximately) satisfying r^2 = b^2 - 4ac, and then the formula z=-b 2a plusorminus r 2a gives us the solutions (as a set): {-b + r 2a, -b-r 2a}. Use this analysis to solve the complex quadratic equation (1 + i)z^2 + 2z - (3 +i) = 0. The solution set is

Explanation / Answer

The given quadratic equation is (1+i)z2 + 2z-(3+i) = 0 . Here a = (1+i), b = 2 and c = -(3+i). Therefore, as per the quadratic formula, we have z = -(b/2a) ± ( r/2a) where r = ( b2 -4ac). Here, r = [ 22 +4*(1+i)(3+i)] = [4+4( 3 + 4i + i2)]   = (12 +16i) ( as i2 = -1). Therefore, z = - 2/2(1+i) ± [(12 + 16i)]/2(1+i) = - (1-i)/(1+i)(1-i) ± [(12 + 16i)](1-i)/2(1+i)(1-i) = -(1-i)/(1-i2)± [(12+16i)( 1-i)2]/2( 1 –i2) = -(1-i)/2 ± ¼ [(12+16i)( 1-2i+i2)] = -(1-i)/2 ± ¼ [(12+16i)(-2i)] = -(1-i)/2 ± ¼ (-24i -32i2) = -(1-i)/2 ± ¼ (32-24i) = -(1-i)/2 ± ½ ( 8-6i).

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