What is (2+2ix-2x)^2 ?

The solution to (2+2ix-2x)^2 is (-8x+4)+(8x-8x^2)i

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simplify (2+2ix-2x)^2

simplify

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

(- 8 x + 4) + (8 x - 8 x^{2}) i

Solution steps

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

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

(2 + 2 i x - 2 x)^{2} = (2 + 2 i x - 2 x) (2 + 2 i x - 2 x)

= (2 + 2 i x - 2 x) (2 + 2 i x - 2 x)

Distribute parentheses

= 2 \cdot 2 + 2 \cdot 2 i x + 2 (- 2 x) + 2 i x \cdot 2 + 2 i x \cdot 2 i x + 2 i x (- 2 x) - 2 x \cdot 2 - 2 x \cdot 2 i x - 2 x (- 2 x)

Simplify

2 \cdot 2 + 2 \cdot 2 i x + 2 (- 2 x) + 2 i x \cdot 2 + 2 i x \cdot 2 i x + 2 i x (- 2 x) - 2 x \cdot 2 - 2 x \cdot 2 i x - 2 x (- 2 x) : 8 x i - 8 x - 8 x^{2} i + 4

= 8 x i - 8 x - 8 x^{2} i + 4

Rewrite in standard complex form:

(- 8 x + 4) + (8 x - 8 x^{2}) i

= (- 8 x + 4) + (8 x - 8 x^{2}) i

e^{x^{2}} (3 x + 5)^{2}

i (1 + 2 e^{i t})

\frac{( - 2 a + 7 b ) ^{2}}{9}

- \frac{18 x}{( 3 x ^{2} + 1 ) ^{2}}

\frac{6 x - 5}{3 ( x - 1 ) ^{\frac{2}{3}} ( 2 x - 1 ) ^{\frac{1}{3}}}