The solution to (x+iy)^4 is (x^4-6x^2y^2+y^4)+(4x^3y-4xy^3)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (x+iy)^4

Solution

simplify

(x + i y)^{4}

(x^{4} - 6 x^{2} y^{2} + y^{4}) + (4 x^{3} y - 4 x y^{3}) i

Solution steps

(x + i y)^{4}

Apply binomial theorem:

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

a = x, b = i y

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

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} x^{4} (i y)^{0} : x^{4}

Simplify

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

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} x^{2} (i y)^{2} : 6 i^{2} x^{2} y^{2}

Simplify

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

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} x^{0} (i y)^{4} : i^{4} y^{4}

= x^{4} + 4 i x^{3} y + 6 i^{2} x^{2} y^{2} + 4 i^{3} x y^{3} + i^{4} y^{4}

6 i^{2} x^{2} y^{2} = - 6 x^{2} y^{2}

4 i^{3} x y^{3} = - 4 i x y^{3}

i^{4} y^{4} = y^{4}

= x^{4} + 4 i x^{3} y - 6 x^{2} y^{2} - 4 i x y^{3} + y^{4}

Rewrite in standard complex form:

(x^{4} - 6 x^{2} y^{2} + y^{4}) + (4 x^{3} y - 4 x y^{3}) i

= (x^{4} - 6 x^{2} y^{2} + y^{4}) + (4 x^{3} y - 4 x y^{3}) i

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

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

cos (x) (csc (x) - sec (x)) - cot (x)

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

(x + y) (x - y)