What is (2+2sqrt(3)i)^4 ?

The solution to (2+2sqrt(3)i)^4 is -128-128sqrt(3)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (2+2sqrt(3)i)^4

simplify

(2 + 23 i)^{4}

- 128 - 1283 i

Solution steps

(2 + 23 i)^{4}

Apply binomial theorem:

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

a = 2, b = 23 i

= i = 0 \sum 4 (i 4) \cdot 2^{(4 - i)} (23 i)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 2^{4} (23 i)^{0} : 16

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 2^{3} (23 i)^{1} : 643 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 2^{2} (23 i)^{2} : 288 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 2^{1} (23 i)^{3} : 1923 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 2^{0} (23 i)^{4} : 144 i^{4}

= 16 + 643 i + 288 i^{2} + 1923 i^{3} + 144 i^{4}

Simplify

16 + 643 i + 288 i^{2} + 1923 i^{3} + 144 i^{4} : - 1283 i - 128

= - 1283 i - 128

Rewrite in standard complex form:

- 128 - 1283 i

= - 128 - 1283 i

f (x, y) 3 y (\frac{x ^{3}}{2} - 1)

\frac{- 4 x + 57}{3 ( 19 - x ) ^{\frac{2}{3}}}

(1 + 2 cos (x))^{2}

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

e^{2 x} (2 x + 1)