The solution to (x+2i)^4 is (x^4-24x^2+16)+(8x^3-32x)i

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simplify (x+2i)^4

Solution

simplify

(x + 2 i)^{4}

(x^{4} - 24 x^{2} + 16) + (8 x^{3} - 32 x) i

Solution steps

(x + 2 i)^{4}

Apply binomial theorem:

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

a = x, b = 2 i

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

Expand summation

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

Simplify

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

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} x^{3} (2 i)^{1} : 8 i x^{3}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} x^{2} (2 i)^{2} : 24 i^{2} x^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} x^{1} (2 i)^{3} : 32 i^{3} x

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} x^{0} (2 i)^{4} : 16 i^{4}

= x^{4} + 8 i x^{3} + 24 i^{2} x^{2} + 32 i^{3} x + 16 i^{4}

24 i^{2} x^{2} = - 24 x^{2}

32 i^{3} x = - 32 i x

16 i^{4} = 16

= x^{4} + 8 i x^{3} - 24 x^{2} - 32 i x + 16

Rewrite in standard complex form:

(x^{4} - 24 x^{2} + 16) + (8 x^{3} - 32 x) i

= (x^{4} - 24 x^{2} + 16) + (8 x^{3} - 32 x) i

10 (x + Δ x)^{2}

(x + 2 i)^{3}

\frac{2 ( x ^{4} - 3 x ^{2} )}{( x ^{2} - 1 ) ^{2}}

\frac{1}{( 3 + n ) ^{2}}