化简 (x+1-i)^2

解答

化简

(x + 1 - i)^{2}

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

求解步骤

(x + 1 - i)^{2}

使用指数法则:

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

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

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

打开括号

= xx + x \cdot 1 + x (- i) + 1 \cdot x + 1 \cdot 1 + 1 \cdot (- i) - i x - i 1 - i (- i)

化简

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

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

Rewrite in standard complex form:

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

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

e x p an d A (5, - 2) B (- 1, 2)

e x p an d (2 c + d + e) (2 c + d - e)

e x p an d \frac{1}{( s - 2 ) ( s ^{2} + 1 )}

e x p an d \frac{( x - 8 ) ^{2}}{( x ^{2} - 16 )}

e x p an d \frac{x ^{2} + 2}{( x + 3 ) ( x + 2 ) ^{2}}