What is (2-sqrt(3)i)^5 ?

The solution to (2-sqrt(3)i)^5 is -118+31sqrt(3)i

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simplify (2-sqrt(3)i)^5

simplify

(2 - 3 i)^{5}

- 118 + 313 i

Solution steps

(2 - 3 i)^{5}

Apply binomial theorem:

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

a = 2, b = - 3 i

= i = 0 \sum 5 (i 5) \cdot 2^{(5 - i)} (- 3 i)^{i}

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 2^{5} (- 3 i)^{0} : 32

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 2^{4} (- 3 i)^{1} : - 803 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 2^{3} (- 3 i)^{2} : 240 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 2^{2} (- 3 i)^{3} : - 1203 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 2^{1} (- 3 i)^{4} : 90 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 2^{0} (- 3 i)^{5} : - 93 i^{5}

= 32 - 803 i + 240 i^{2} - 1203 i^{3} + 90 i^{4} - 93 i^{5}

Simplify

32 - 803 i + 240 i^{2} - 1203 i^{3} + 90 i^{4} - 93 i^{5} : 313 i - 118

= 313 i - 118

Rewrite in standard complex form:

- 118 + 313 i

= - 118 + 313 i

(\frac{3 a}{2 b} + \frac{4 b ^{2}}{5})^{3}

(x - \frac{2}{x})^{10}

\frac{d ^{2}}{d x ^{2}} ((y) (x^{2} + 2 x) (3 x - 2)^{2})

(4, 2, - 3, 5, - 1) l e f t (2, 6, - 1, - 4, 8)

(2 x - 5)^{2}