What is the answer to (z+1)^2=3+4i ?

The answer to (z+1)^2=3+4i is z=1+i,z=-3-i

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(z+1)^2=3+4i

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

z = 1 + i, z = - 3 - i

Solution steps

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

Substitute

z = x + y i

(x + y i + 1)^{2} = 3 + 4 i

Expand

(x + y i + 1)^{2} : (x^{2} + 2 x - y^{2} + 1) + i (2 y + 2 x y)

(x^{2} + 2 x - y^{2} + 1) + i (2 y + 2 x y) = 3 + 4 i

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

[x^{2} + 2 x - y^{2} + 1 = 3 2 y + 2 x y = 4]

[x^{2} + 2 x - y^{2} + 1 = 3 2 y + 2 x y = 4] : (x = 1, x = - 3, y = 1 y = - 1)

Substitute back

z = x + y i

z = 1 + i, z = - 3 - i

7 x + x i - 3 y i = 3 y + 9 i

z^{2} = - i

z^{2} = - 8 i

x - 2 i = 3 + 3 i - y i

A = P (1 + \frac{i}{100})^{2}