What is the answer to |z+3|=1-iz ?

The answer to |z+3|=1-iz is z=4i

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|z+3|=1-iz

∣ z + 3 ∣ = 1 - i z

z = 4 i

Solution steps

∣ z + 3 ∣ = 1 - i z

Substitute

z = x + y i

∣ x + y i + 3 ∣ = 1 - i (x + y i)

Expand

x^{2} + 6 x + 9 + y^{2} = (1 + y) - i x

Rewrite

x^{2} + 6 x + 9 + y^{2}

in standard complex form:

x^{2} + 6 x + 9 + y^{2} + 0 i

x^{2} + 6 x + 9 + y^{2} + 0 i = (1 + y) - i x

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

[x^{2} + 6 x + 9 + y^{2} = 1 + y 0 = - x]

[x^{2} + 6 x + 9 + y^{2} = 1 + y 0 = - x] : (x = 0, y = 4)

Substitute back

z = x + y i

z = 4 i

2 x + 8 i = 6 - (2 y) i

9 x + (y) i - 5 = - 12 i + 4

i = (a + bi)^{2}

\frac{1 + ai}{5 + bi} = 2 + 7 i

z - \frac{2}{z} = i + \frac{2}{i}