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

The solution to (-sqrt(3)+i)^5 is 16sqrt(3)+16i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

simplify

(- 3 + i)^{5}

163 + 16 i

Solution steps

(- 3 + i)^{5}

Apply binomial theorem:

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

a = - 3, b = i

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

Expand summation

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

Simplify

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

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (- 3)^{4} i^{1} : 45 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (- 3)^{3} i^{2} : - 303 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (- 3)^{2} i^{3} : 30 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (- 3)^{1} i^{4} : - 53 i^{4}

Simplify

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

= - 93 + 45 i - 303 i^{2} + 30 i^{3} - 53 i^{4} + i^{5}

Simplify

- 93 + 45 i - 303 i^{2} + 30 i^{3} - 53 i^{4} + i^{5} : 16 i + 163

= 16 i + 163

Rewrite in standard complex form:

163 + 16 i

= 163 + 16 i

x (x + 1)

(2 + 3 - 2 - 3)^{6}

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

(x - 4)^{2}

(sec (x) + tan (x) - 1) (sec (x) - tan (x) + 1)