The solution to (a-bi)^4 is (a^4-6a^2b^2+b^4)+(-4a^3b+4ab^3)i

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simplify (a-bi)^4

Solution

simplify

(a - bi)^{4}

(a^{4} - 6 a^{2} b^{2} + b^{4}) + (- 4 a^{3} b + 4 a b^{3}) i

Solution steps

(a - bi)^{4}

Apply binomial theorem:

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

a = a, b = - bi

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

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} a^{4} (- bi)^{0} : a^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} a^{3} (- bi)^{1} : - 4 i a^{3} b

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} a^{2} (- bi)^{2} : 6 i^{2} a^{2} b^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} a^{1} (- bi)^{3} : - 4 i^{3} a b^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} a^{0} (- bi)^{4} : i^{4} b^{4}

= a^{4} - 4 i a^{3} b + 6 i^{2} a^{2} b^{2} - 4 i^{3} a b^{3} + i^{4} b^{4}

Simplify

a^{4} - 4 i a^{3} b + 6 i^{2} a^{2} b^{2} - 4 i^{3} a b^{3} + i^{4} b^{4} : a^{4} - 4 i a^{3} b - 6 a^{2} b^{2} + 4 ia b^{3} + b^{4}

= a^{4} - 4 i a^{3} b - 6 a^{2} b^{2} + 4 ia b^{3} + b^{4}

Rewrite in standard complex form:

(a^{4} - 6 a^{2} b^{2} + b^{4}) + (- 4 a^{3} b + 4 a b^{3}) i

= (a^{4} - 6 a^{2} b^{2} + b^{4}) + (- 4 a^{3} b + 4 a b^{3}) i

(x + 8)^{2}

(- 2, 7) y (2, - 1)

\frac{\partial}{\partial x} ((z) (3 x^{2} y + 4 x y^{2}))

6 (11 + 5)

\frac{( x - 1 ) ^{4}}{4}

What is (a-bi)^4 ?
The solution to (a-bi)^4 is (a^4-6a^2b^2+b^4)+(-4a^3b+4ab^3)i