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

The answer to -1+2i=(2i)/((z-1)) is z= 9/5-2/5 i

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-1+2i=(2i)/((z-1))

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

z = \frac{9}{5} - \frac{2}{5} i

Solution steps

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

Substitute

z = x + y i

- 1 + 2 i = \frac{2 i}{x + y i - 1}

Multiply both sides by

x - 1 + i y

- 1 \cdot (x - 1 + i y) + 2 i (x - 1 + i y) = \frac{2 i}{x + y i - 1} (x - 1 + i y)

Expand

(- x + 1 - 2 y) + i (- 2 - y + 2 x) = 2 i

Rewrite

2 i

in standard complex form:

0 + 2 i

(- x + 1 - 2 y) + i (- 2 - y + 2 x) = 0 + 2 i

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

[- x + 1 - 2 y = 0 - 2 - y + 2 x = 2]

[- x + 1 - 2 y = 0 - 2 - y + 2 x = 2] : x = \frac{9}{5}, y = - \frac{2}{5}

Substitute back

z = x + y i

z = \frac{9}{5} - \frac{2}{5} i

- x + y i + 7 = 6 i

0 = z^{2} + i z - 6

(7 + m) + (4 l - 10) i = 3 - 6 i

(- 2 - 6 i) - p (4 - 2 i) = q i

(x - y i) + (y + x i) = (5 + i)

What is the answer to -1+2i=(2i)/((z-1)) ?
The answer to -1+2i=(2i)/((z-1)) is z= 9/5-2/5 i