What is the answer to i(t)=8t^2-4tA ?

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i(t)=8t^2-4tA

i (t) = 8 t^{2} - 4 t A

(t = 0, A \in R)

Solution steps

i (t) = 8 t^{2} - 4 t A

Rewrite

i (t)

in standard complex form:

0 + t i

0 + t i = 8 t^{2} - 4 t A

Rewrite

8 t^{2} - 4 t A

in standard complex form:

(8 t^{2} - 4 t A) + 0 i

0 + t i = (8 t^{2} - 4 t A) + 0 i

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

[0 = 8 t^{2} - 4 t A t = 0]

[0 = 8 t^{2} - 4 t A t = 0] : (t = 0, A \in R)

(t = 0, A \in R)

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

∣ z + i ∣ = 3 + 0 i

x^{2} - 3 (1 - i) x - 5 i = 0

\frac{( x - 1 )}{2 - x} = i

x y + (x - y) i = 4