Solving equations with complex numbers with two variables In this section we will solve equations with complex numbers involving two variables. Example 1: If (x + = determine the values of and if . yi)(2 i) x y x, y – –i, Solution: Method 1: (x + = yi)(2 i) – –i, ( = both sides by x + yi) (2 i)] ………..[Divide – (x = RHS by conjugate of denominator] + yi) ……..[Multiply (x = + yi) (x = = + yi) i2 ……..[ –1] ( = numerator and denominator] x + yi) ………...[Simplify = + x + yi = x + yi – Now, comparing real part and imaginary parts of the equation: = and = x y – Method 2: (x + = yi)(2 i) – –i, 2x + 2yi = LHS} xi yi2 i – – – ……[Expand 2x + 2yi = = y(–1) xi i – – – ……[i2 –1] 2x + (x = real terms and imaginary terms] y i – –2y)i – ……..[Group 2x + (x 2y)i = 0 RHS in form y i…….[Write a + bi] – – – Comparing real parts: 2 + = 0 x y ………..(1) Comparing imaginary parts: 2y = 1……..(2) x – Solving the above equations simultaneously as follows: (2) 2: 2x = 2 –4y …….(3) (3) (1): = 2 – –5y = y – Substitute = in (1): 2x = 0 y –
Solving equations with complex numbers with two variables In this section we will solve equations with complex numbers involving two variables. Example 1: If (x + yi)(2 – i) = –i, determine the values of x and y if x, y . Solution: Method 1: (x + yi)(2 – i) = –i, ( x + yi) = ………..[Divide both sides by (2 – i)] (x + yi) = ……..[Multiply RHS by conjugate of denominator] (x + yi) = (x + yi) = ……..[ i2 = –1] ( x + yi) = ………...[Simplify numerator and denominator] x + yi = + x + yi = – Now, comparing real part and imaginary parts of the equation: x = and y = – Method 2: (x + yi)(2 – i) = –i, 2x – xi + 2yi – yi2 = – i ……[Expand LHS} 2x – y(–1) – xi + 2yi = – i ……[i2 = –1] 2x + y – (x –2y)i = – i ……..[Group real terms and imaginary terms] 2x + y – (x – 2y)i = 0 – i…….[Write RHS in form a + bi] Comparing real parts: 2 x + y = 0 ………..(1) Comparing imaginary parts: x – 2y = 1……..(2) Solving the above equations simultaneously as follows: (2) 2: 2x –4y = 2 …….(3) (3) – (1): –5y = 2 y = – Substitute y = in (1): 2x – = 0
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