What is the answer to -3-4i=(a+bi)^2 ?

The answer to -3-4i=(a+bi)^2 is

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-3-4i=(a+bi)^2

- 3 - 4 i = (a + bi)^{2}

(a = - 1, a = 1, b = 2 b = - 2)

Solution steps

- 3 - 4 i = (a + bi)^{2}

Expand

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

- 3 - 4 i = (a^{2} - b^{2}) + 2 iab

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

[- 3 = a^{2} - b^{2} - 4 = 2 ab]

[- 3 = a^{2} - b^{2} - 4 = 2 ab] : (a = - 1, a = 1, b = 2 b = - 2)

(a = - 1, a = 1, b = 2 b = - 2)

5 x + (7 x - 2 y) i = 2 + 8 i

(3 + 4 i) (u + v i) = 5 - i

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

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

z^{2} - z (1 + 4 i) + 3 + 11 i = 0