What is the answer to a+bi-1/(a+bi)=2+4i ?

The answer to a+bi-1/(a+bi)=2+4i is

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a+bi-1/(a+bi)=2+4i

a + bi - \frac{1}{a + bi} = 2 + 4 i

(a = - 0.11178 \dots, a = 2.11178 \dots, b = 0.20109 \dots b = 3.79890 \dots)

Solution steps

a + bi - \frac{1}{a + bi} = 2 + 4 i

Multiply both sides by

a + bi

a (a + bi) + bi (a + bi) - \frac{1}{a + bi} (a + bi) = 2 (a + bi) + 4 i (a + bi)

Expand

(a^{2} - b^{2} - 1) + 2 iab = (2 a - 4 b) + i (2 b + 4 a)

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[a^{2} - b^{2} - 1 = 2 a - 4 b 2 ab = 2 b + 4 a]

[a^{2} - b^{2} - 1 = 2 a - 4 b 2 ab = 2 b + 4 a] : (a = - 0.11178 \dots, a = 2.11178 \dots, b = 0.20109 \dots b = 3.79890 \dots)

(a = - 0.11178 \dots, a = 2.11178 \dots, b = 0.20109 \dots b = 3.79890 \dots)

(a - bi) = i (a + bi)

b^{2} = 3^{2} + (3 i)^{2}

- \frac{1}{z - i} + 1 - z = 0

+ 5 i z^{2} = - 2 + 3 i, z^{1} = 4

z^{1} + z^{2} = (4 + 3) + (2 + 5) i