Finding the square root of a complex number

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Presentation transcript:

Finding the square root of a complex number (a+bi)2 = ( a2 + b2) +2abi so -2i = a + bi Find -2i Let -2i = ( a+bi)2 But a and b are REAL -2i =( a2 + b2) +2abi Equate Real Parts: 0 = a2 + b2 a = -1/b Equate Im Parts: -2 = 2ab 0 = (-1/b)2 + b2 0 = 1/b2 + b2 0 = 1 + b4 (b2 - 1 ) ( b2 + 1 ) = 0 b2 =  1 b = +1, -1, or i

Solve the following equation (a+bi)2 = ( a2 + b2) +2abi Solve z2 + 4z +(4+2i)=0 Z = -4   42 – [4x1x(4+2i)] 2 Z = -4   16 – 16 - 8i 2 Z = -4   – 8i =-2 -2i 2 -2i =( a2 + b2) +2abi so b4 – 26b2 +16 = 0 Equate Real Parts: 0= a2 + b2 Equate Im Parts: -2 = 2ab So b = +1, -1, or i 0 = (-1/b)2 + b2 a = -1/b 0 = 1/b2 + b2 So a = -1, +1, or i 0 = 1 + b4 -3 + i Or -1 - i (b2 - 1 ) ( b2 + 1 ) = 0 So z = -2 -1+i = or -2+1 –i = b2 =  1