What is the answer to (a+i)^2=8+bi ?

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(a+i)^2=8+bi

(a + i)^{2} = 8 + bi

(a = - 3, a = 3, b = - 6 b = 6)

Solution steps

(a + i)^{2} = 8 + bi

Expand

(a + i)^{2} : (a^{2} - 1) + 2 ia

(a^{2} - 1) + 2 ia = 8 + bi

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

[a^{2} - 1 = 8 2 a = b]

[a^{2} - 1 = 8 2 a = b] : (a = - 3, a = 3, b = - 6 b = 6)

(a = - 3, a = 3, b = - 6 b = 6)

\frac{3 + ai}{5 + bi} = 3 + 5 i

z^{2} + (- 3 + 2 i) z + (5 - i) = 0

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

z^{2} = 3 + 3 i

z^{2} = i (1 - 0. 5^{2})