What is expand (1.6707100000000…+0.8976700000000…*i)^4 ?

The solution to expand (1.6707100000000…+0.8976700000000…*i)^4 is -5.05490…+11.91074…i

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expand (1.6707100000000…+0.8976700000000…*i)^4

expand

(1.6707100000000 \dots + 0.8976700000000 \dots \cdot i)^{4}

- 5.05490 \dots + 11.91074 \dots i

Solution steps

(1.67071 \dots + 0.89767 \dots i)^{4}

Apply binomial theorem:

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

a = 1.67071 \dots, b = 0.89767 \dots i

= i = 0 \sum 4 (i 4) \cdot 1.67071 \dots^{(4 - i)} (0.89767 \dots i)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1.67071 \dots^{4} (0.89767 \dots i)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1.67071 \dots^{3} (0.89767 \dots i)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1.67071 \dots^{2} (0.89767 \dots i)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1.67071 \dots^{1} (0.89767 \dots i)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1.67071 \dots^{0} (0.89767 \dots i)^{4}

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1.67071 \dots^{4} (0.89767 \dots i)^{0} = 7.79119 \dots

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1.67071 \dots^{3} (0.89767 \dots i)^{1} : 16.74479 \dots i

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1.67071 \dots^{2} (0.89767 \dots i)^{2} = - 13.49543 \dots

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1.67071 \dots^{1} (0.89767 \dots i)^{3} : - 4.83405 \dots i

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1.67071 \dots^{0} (0.89767 \dots i)^{4} = 0.64933 \dots

= 7.79119 \dots + 16.74479 \dots i - 13.49543 \dots - 4.83405 \dots i + 0.64933 \dots

Simplify

7.79119 \dots + 16.74479 \dots i - 13.49543 \dots - 4.83405 \dots i + 0.64933 \dots : - 5.05490 \dots + 11.91074 \dots i

= - 5.05490 \dots + 11.91074 \dots i

\frac{3 s ^{2} + e ^{- 4 s}}{s ( s ^{2} + 1 )}

(4 - 3 i) + (- 1, 5) - (8, - 4)

(A + (C \cdot cos (y)) + (D \cdot cos (z)) - (B \cdot cos (x)))^{2}

\frac{2 x + 3}{2 ( x + 1 )}

2 (1, 2) + 3 (- 3, 4)^{'} + 2 (- 1, - 2)

What is expand (1.6707100000000…+0.8976700000000…*i)^4 ?
The solution to expand (1.6707100000000…+0.8976700000000…*i)^4 is -5.05490…+11.91074…i