The solution to (a+bi)^5 is (a^5-10a^3b^2+5ab^4)+(5a^4b-10a^2b^3+b^5)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (a+bi)^5

Solution

simplify

(a + bi)^{5}

(a^{5} - 10 a^{3} b^{2} + 5 a b^{4}) + (5 a^{4} b - 10 a^{2} b^{3} + b^{5}) i

Solution steps

(a + bi)^{5}

Apply binomial theorem:

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

a = a, b = bi

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

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} a^{5} (bi)^{0} : a^{5}

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} a^{4} (bi)^{1} : 5 i a^{4} b

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} a^{3} (bi)^{2} : 10 i^{2} a^{3} b^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} a^{2} (bi)^{3} : 10 i^{3} a^{2} b^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} a^{1} (bi)^{4} : 5 i^{4} a b^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} a^{0} (bi)^{5} : i^{5} b^{5}

= a^{5} + 5 i a^{4} b + 10 i^{2} a^{3} b^{2} + 10 i^{3} a^{2} b^{3} + 5 i^{4} a b^{4} + i^{5} b^{5}

10 i^{2} a^{3} b^{2} = - 10 a^{3} b^{2}

10 i^{3} a^{2} b^{3} = - 10 i a^{2} b^{3}

5 i^{4} a b^{4} = 5 a b^{4}

i^{5} = i

= a^{5} + 5 i a^{4} b - 10 a^{3} b^{2} - 10 i a^{2} b^{3} + 5 a b^{4} + i b^{5}

Rewrite in standard complex form:

(a^{5} - 10 a^{3} b^{2} + 5 a b^{4}) + (5 a^{4} b - 10 a^{2} b^{3} + b^{5}) i

= (a^{5} - 10 a^{3} b^{2} + 5 a b^{4}) + (5 a^{4} b - 10 a^{2} b^{3} + b^{5}) i

2 (3 e^{- x} x^{2} - e^{- x} x^{3})

cos (7 t) (5 + sinh (3 t) + e^{2 t})

\frac{d}{d t} ((h (t)) (2.8 t + 2.5))

(4 + 5 y i)^{2}

x^{\frac{4}{5}} (x - 2)^{2}

What is (a+bi)^5 ?
The solution to (a+bi)^5 is (a^5-10a^3b^2+5ab^4)+(5a^4b-10a^2b^3+b^5)i