The solution to (x+1+i)^4 is (x^4+4x^3-8x-4)+(4x^3+8x+12x^2)i

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simplify (x+1+i)^4

Solution

simplify

(x + 1 + i)^{4}

(x^{4} + 4 x^{3} - 8 x - 4) + (4 x^{3} + 8 x + 12 x^{2}) i

Solution steps

(x + 1 + i)^{4}

Apply exponent rule:

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

(x + 1 + i)^{4} = (x + 1 + i) (x + 1 + i)^{3}

= (x + 1 + i) (x + 1 + i)^{3}

Apply exponent rule:

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

(x + 1 + i)^{3} = (x + 1 + i) (x + 1 + i)^{2}

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

Apply exponent rule:

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

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

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

Expand

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

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

Expand

(2 i + x^{2} + 2 x + 2 x i) (x + 1 + i) : 2 i + x^{3} + 3 x^{2} + 3 x^{2} i + 6 x i - 2

= (2 i + x^{3} + 3 x^{2} + 3 x^{2} i + 6 x i - 2) (x + 1 + i)

Distribute parentheses

= 2 i x + 2 i 1 + 2 ii + x^{3} x + x^{3} \cdot 1 + x^{3} i + 3 x^{2} x + 3 x^{2} \cdot 1 + 3 x^{2} i + 3 x^{2} i x + 3 x^{2} i 1 + 3 x^{2} ii + 6 x i x + 6 x i 1 + 6 x ii - 2 x - 2 \cdot 1 - 2 i

Simplify

2 i x + 2 i 1 + 2 ii + x^{3} x + x^{3} \cdot 1 + x^{3} i + 3 x^{2} x + 3 x^{2} \cdot 1 + 3 x^{2} i + 3 x^{2} i x + 3 x^{2} i 1 + 3 x^{2} ii + 6 x i x + 6 x i 1 + 6 x ii - 2 x - 2 \cdot 1 - 2 i : x^{4} + 4 x^{3} + 4 x^{3} i + 8 x i + 12 x^{2} i - 8 x - 4

= x^{4} + 4 x^{3} + 4 x^{3} i + 8 x i + 12 x^{2} i - 8 x - 4

Rewrite in standard complex form:

(x^{4} + 4 x^{3} - 8 x - 4) + (4 x^{3} + 8 x + 12 x^{2}) i

= (x^{4} + 4 x^{3} - 8 x - 4) + (4 x^{3} + 8 x + 12 x^{2}) i

\frac{3 ( - \frac{7}{8} ) ^{n} + 5}{17}

(x^{4} - 9) \frac{d y}{d x} + x^{3} y

x - 3 (3 (x + h)^{2} - 2 x + 2 h)

\frac{1}{( 1 + x ^{2} ) \cdot ( x ^{2} + x + 1 )}

\frac{24 ( s + 9 )}{( s + 0.8 ) ( s + 27 )}

What is (x+1+i)^4 ?
The solution to (x+1+i)^4 is (x^4+4x^3-8x-4)+(4x^3+8x+12x^2)i