The solution to (x+i)^4 is (x^4-6x^2+1)+(4x^3-4x)i

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simplify (x+i)^4

Solution

simplify

(x + i)^{4}

(x^{4} - 6 x^{2} + 1) + (4 x^{3} - 4 x) i

Solution steps

(x + i)^{4}

Apply binomial theorem:

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

a = x, b = i

= i = 0 \sum 4 (i 4) x^{(4 - i)} i^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} x^{4} i^{0} : x^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} x^{3} i^{1} : 4 i x^{3}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} x^{2} i^{2} : 6 i^{2} x^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} x^{1} i^{3} : 4 i^{3} x

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} x^{0} i^{4} : i^{4}

= x^{4} + 4 i x^{3} + 6 i^{2} x^{2} + 4 i^{3} x + i^{4}

6 i^{2} x^{2} = - 6 x^{2}

4 i^{3} x = - 4 i x

i^{4} = 1

= x^{4} + 4 i x^{3} - 6 x^{2} - 4 i x + 1

Rewrite in standard complex form:

(x^{4} - 6 x^{2} + 1) + (4 x^{3} - 4 x) i

= (x^{4} - 6 x^{2} + 1) + (4 x^{3} - 4 x) i

(8 + 7) (9 + 2)

(x + 5 x^{2})^{2}

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

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

(\frac{k ( k + 1 )}{2})^{2}