What is (x+i(-x+1))^2 ?

The solution to (x+i(-x+1))^2 is (2x-1)+(-2x^2+2x)i

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simplify (x+i(-x+1))^2

simplify

(x + i (- x + 1))^{2}

(2 x - 1) + (- 2 x^{2} + 2 x) i

Solution steps

(x + i (- x + 1))^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(x + i (- x + 1))^{2} = x^{2} + 2 x i (- x + 1) + (i (- x + 1))^{2}

= x^{2} + 2 x i (- x + 1) + (i (- x + 1))^{2}

Simplify

x^{2} + 2 x i (- x + 1) + (i (- x + 1))^{2} : 2 x i (- x + 1) + 2 x - 1

= 2 x i (- x + 1) + 2 x - 1

Expand

2 x i (- x + 1) : - 2 x^{2} i + 2 x i

= - 2 x^{2} i + 2 x i + 2 x - 1

Rewrite in standard complex form:

(2 x - 1) + (- 2 x^{2} + 2 x) i

= (2 x - 1) + (- 2 x^{2} + 2 x) i

(5 + 23)^{2} - 10 (5 + 23) + 13

p^{(n - m - 1)^{2}}

\frac{lo g _{10} ( 3 ) [ ( x + y ) ^{2} ]}{[ x ^{2} + y ^{2} ]}

\frac{d}{d M} ((R) (M^{2} (\frac{c}{2} - \frac{M}{3})))

\frac{( 3 s + 8 )}{( s + 3 ) ^{2}}