What is (7-3i)+(x-2i)^2-(4i+2x^2) ?

The solution to (7-3i)+(x-2i)^2-(4i+2x^2) is (-x^2+3)+i(-7-4x)

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simplify (7-3i)+(x-2i)^2-(4i+2x^2)

simplify

(7 - 3 i) + (x - 2 i)^{2} - (4 i + 2 x^{2})

(- x^{2} + 3) + i (- 7 - 4 x)

Solution steps

(7 - 3 i) + (x - 2 i)^{2} - (4 i + 2 x^{2})

Remove parentheses:

(a) = a

= 7 - 3 i + (x - 2 i)^{2} - (4 i + 2 x^{2})

(x - 2 i)^{2} : x^{2} - 4 i x - 4

= 7 - 3 i + x^{2} - 4 i x - 4 - (4 i + 2 x^{2})

- (4 i + 2 x^{2}) : - 4 i - 2 x^{2}

= 7 - 3 i + x^{2} - 4 i x - 4 - 4 i - 2 x^{2}

Simplify

7 - 3 i + x^{2} - 4 i x - 4 - 4 i - 2 x^{2} : - x^{2} - 4 i x + 3 - 7 i

= - x^{2} - 4 i x + 3 - 7 i

Rewrite

- x^{2} - 4 i x + 3 - 7 i

in standard complex form:

(- x^{2} + 3) + (- 4 x - 7) i

= (- x^{2} + 3) + (- 4 x - 7) i

cot (x) tan (x)

9 x^{2} - 39 x - 30

\frac{x ^{3}}{2} (4 x^{- 1})^{2}

\frac{1}{2} (2 x^{2} - 8)^{2} + 8

a x^{2} + b x

What is (7-3i)+(x-2i)^2-(4i+2x^2) ?
The solution to (7-3i)+(x-2i)^2-(4i+2x^2) is (-x^2+3)+i(-7-4x)