What is (sqrt(3)-3i)^4 ?

The solution to (sqrt(3)-3i)^4 is -72+72sqrt(3)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (sqrt(3)-3i)^4

simplify

(3 - 3 i)^{4}

- 72 + 723 i

Solution steps

(3 - 3 i)^{4}

Apply binomial theorem:

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

a = 3, b = - 3 i

= i = 0 \sum 4 (i 4) (3)^{(4 - i)} (- 3 i)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (3)^{4} (- 3 i)^{0} : 9

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (3)^{3} (- 3 i)^{1} : - 363 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (3)^{2} (- 3 i)^{2} : 162 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (3)^{1} (- 3 i)^{3} : - 1083 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (3)^{0} (- 3 i)^{4} : 81 i^{4}

= 9 - 363 i + 162 i^{2} - 1083 i^{3} + 81 i^{4}

Simplify

9 - 363 i + 162 i^{2} - 1083 i^{3} + 81 i^{4} : 723 i - 72

= 723 i - 72

Rewrite in standard complex form:

- 72 + 723 i

= - 72 + 723 i

\frac{( x - 2 ) ^{2}}{25}

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

\frac{20 e ^{- 4 s}}{s ( s ^{2} + 4 s + 20 )}

x (6 + 5 x)

(1 + 6 sin (θ))^{2}