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

The solution to (2sqrt(3)+2i)^5 is -512sqrt(3)+512i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

simplify

(23 + 2 i)^{5}

- 5123 + 512 i

Solution steps

(23 + 2 i)^{5}

Apply binomial theorem:

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

a = 23, b = 2 i

= i = 0 \sum 5 (i 5) (23)^{(5 - i)} (2 i)^{i}

Expand summation

= \frac{5 !}{0 ! ( 5 - 0 ) !} (23)^{5} (2 i)^{0} + \frac{5 !}{1 ! ( 5 - 1 ) !} (23)^{4} (2 i)^{1} + \frac{5 !}{2 ! ( 5 - 2 ) !} (23)^{3} (2 i)^{2} + \frac{5 !}{3 ! ( 5 - 3 ) !} (23)^{2} (2 i)^{3} + \frac{5 !}{4 ! ( 5 - 4 ) !} (23)^{1} (2 i)^{4} + \frac{5 !}{5 ! ( 5 - 5 ) !} (23)^{0} (2 i)^{5}

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} (23)^{5} (2 i)^{0} : 2883

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (23)^{4} (2 i)^{1} : 1440 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (23)^{3} (2 i)^{2} : 9603 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (23)^{2} (2 i)^{3} : 960 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (23)^{1} (2 i)^{4} : 1603 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} (23)^{0} (2 i)^{5} : 32 i^{5}

= 2883 + 1440 i + 9603 i^{2} + 960 i^{3} + 1603 i^{4} + 32 i^{5}

Simplify

2883 + 1440 i + 9603 i^{2} + 960 i^{3} + 1603 i^{4} + 32 i^{5} : 512 i - 5123

= 512 i - 5123

Rewrite in standard complex form:

- 5123 + 512 i

= - 5123 + 512 i

\frac{3}{s ( s ^{2} + 2 s + 5 )}

(3 e + 4) (3 e - 4)

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

(x + 23) (x - 23)

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