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受欢迎的代数 >

化简 (1+sqrt(2)i)^4

解答

化简

(1 + 2 i)^{4}

解答

- 7 - 42 i

求解步骤

(1 + 2 i)^{4}

使用二项式定理:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 1, b = 2 i

= i = 0 \sum 4 (i 4) \cdot 1^{(4 - i)} (2 i)^{i}

展开求和

= \frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1^{4} (2 i)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1^{3} (2 i)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (2 i)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (2 i)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (2 i)^{4}

化简

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1^{4} (2 i)^{0} : 1

化简

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1^{3} (2 i)^{1} : 42 i

化简

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (2 i)^{2} : 12 i^{2}

化简

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (2 i)^{3} : 82 i^{3}

化简

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (2 i)^{4} : 4 i^{4}

= 1 + 42 i + 12 i^{2} + 82 i^{3} + 4 i^{4}

化简

1 + 42 i + 12 i^{2} + 82 i^{3} + 4 i^{4} : - 42 i - 7

= - 42 i - 7

Rewrite in standard complex form:

- 7 - 42 i

= - 7 - 42 i

流行的例子

展开 (49sqrt(7)(-7pi+60))/(15)

e x p an d \frac{49 7 ( - 7 π + 60 )}{15}

展开 (4x(x^2-28x-196)(7+x))/(147)

e x p an d \frac{4 x ( x ^{2} - 28 x - 196 ) ( 7 + x )}{147}

展开 (1-cos(x)-sin(x))^2

e x p an d (1 - cos (x) - sin (x))^{2}

展开 (x^2)/((x-1)(x+1))

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

展开 (x-4)/(2(x+4))

e x p an d \frac{x - 4}{2 ( x + 4 )}