en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (2-2i)^5

Solution

simplify

(2 - 2 i)^{5}

Solution

- 128 + 128 i

Solution steps

(2 - 2 i)^{5}

Apply binomial theorem:

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

a = 2, b = - 2 i

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

Expand summation

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

Simplify

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

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 2^{4} (- 2 i)^{1} : - 160 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 2^{3} (- 2 i)^{2} : 320 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 2^{2} (- 2 i)^{3} : - 320 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 2^{1} (- 2 i)^{4} : 160 i^{4}

Simplify

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

= 32 - 160 i + 320 i^{2} - 320 i^{3} + 160 i^{4} - 32 i^{5}

Simplify

32 - 160 i + 320 i^{2} - 320 i^{3} + 160 i^{4} - 32 i^{5} : 128 i - 128

= 128 i - 128

Rewrite in standard complex form:

- 128 + 128 i

= - 128 + 128 i

Popular Examples

expand (3+3sin(x))^2 expand

(3 + 3 sin (x))^{2}

expand (2x+3)/((x-1)(x^2+4))expand

\frac{2 x + 3}{( x - 1 ) ( x ^{2} + 4 )}

expand 1/((x-1)(x+1))expand

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

expand (2x)/((1-x^2)^2)expand

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

simplify (3+3i)^7 simplify

(3 + 3 i)^{7}

Frequently Asked Questions (FAQ)

What is (2-2i)^5 ?
The solution to (2-2i)^5 is -128+128i