What is (1+sqrt(2)i)^5 ?

The solution to (1+sqrt(2)i)^5 is 1-11sqrt(2)i

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simplify (1+sqrt(2)i)^5

simplify

(1 + 2 i)^{5}

1 - 112 i

Solution steps

(1 + 2 i)^{5}

Apply binomial theorem:

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

a = 1, b = 2 i

= i = 0 \sum 5 (i 5) \cdot 1^{(5 - i)} (2 i)^{i}

Expand summation

= \frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 1^{5} (2 i)^{0} + \frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 1^{4} (2 i)^{1} + \frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 1^{3} (2 i)^{2} + \frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 1^{2} (2 i)^{3} + \frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 1^{1} (2 i)^{4} + \frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 1^{0} (2 i)^{5}

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 1^{5} (2 i)^{0} : 1

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 1^{4} (2 i)^{1} : 52 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 1^{3} (2 i)^{2} : 20 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 1^{2} (2 i)^{3} : 202 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 1^{1} (2 i)^{4} : 20 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 1^{0} (2 i)^{5} : 42 i^{5}

= 1 + 52 i + 20 i^{2} + 202 i^{3} + 20 i^{4} + 42 i^{5}

Simplify

1 + 52 i + 20 i^{2} + 202 i^{3} + 20 i^{4} + 42 i^{5} : - 112 i + 1

= - 112 i + 1

Rewrite in standard complex form:

1 - 112 i

= 1 - 112 i

6 x (3 - 2 x)

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

\frac{x ^{5}}{( x ^{2} + 4 ) ^{2}}

x (1 + x)

(tan (x) - 2) (tan (x) + 2)