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

The answer to (1+(a+bi))/(1-(a+bi))=2i is a= 3/5 ,b= 4/5

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

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

a = \frac{3}{5}, b = \frac{4}{5}

Solution steps

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

Multiply both sides by

1 - a - ib

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

Expand

(1 + a) + ib = 2 b + 2 i (1 - a)

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

[1 + a = 2 b b = 2 - 2 a]

[1 + a = 2 b b = 2 - 2 a] : a = \frac{3}{5}, b = \frac{4}{5}

a = \frac{3}{5}, b = \frac{4}{5}

x^{2} + i x - 1 = 0

(x + i y)^{2} = 3 - 4 i

z^{2} + i z + 1 = 0

\frac{9}{2} x + (3 i) y = 0

y, (- 2 + 5 i) (1 - i) + (4 + 3 i) = - 6 x + 2 y - 3

What is the answer to (1+(a+bi))/(1-(a+bi))=2i ?
The answer to (1+(a+bi))/(1-(a+bi))=2i is a= 3/5 ,b= 4/5