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

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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}

Simplify

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

\frac{- 6}{2 x - 1}

(2 sin (x) + 1) (sin (x) + 1)

cos^{2} (t) - sin^{2} (t) u (t - π)

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

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

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