What is the answer to |z+1-i|=1 ?

The answer to |z+1-i|=1 is z=x+(1+sqrt(-x(x+2)))i,z=x+(1-sqrt(-x(x+2)))i;\quad x\in \mathbb{R}

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|z+1-i|=1

∣ z + 1 - i ∣ = 1

z = x + (1 + - x (x + 2)) i, z = x + (1 - - x (x + 2)) i; x \in R

Solution steps

∣ z + 1 - i ∣ = 1

Substitute

z = x + y i

∣ x + y i + 1 - i ∣ = 1

Expand

∣ x + y i + 1 - i ∣ : x^{2} + 2 x + y^{2} + 2 - 2 y

x^{2} + 2 x + y^{2} + 2 - 2 y = 1

Solve for

y : y = 1 + - x (x + 2), y = 1 - - x (x + 2)

Substitute back

z = x + y i

z = x + (1 + - x (x + 2)) i, z = x + (1 - - x (x + 2)) i; x \in R

\frac{2 - 2 i}{4} = \frac{1 - 2 i}{z}

i z + (1 + 2 i) w = 25

x^{2} + 6 + (2 x) i = 15 + 6 i

i = \frac{70 \cdot 10 e ^{- 9}}{5 s}

e = \frac{( i \cdot n ^{2} )}{( 179 )}

What is the answer to |z+1-i|=1 ?
The answer to |z+1-i|=1 is z=x+(1+sqrt(-x(x+2)))i,z=x+(1-sqrt(-x(x+2)))i;\quad x\in \mathbb{R}