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

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simplify (a+ib)^4

Solution

simplify

(a + ib)^{4}

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

Solution steps

(a + ib)^{4}

Apply binomial theorem:

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

a = a, b = ib

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

6 i^{2} a^{2} b^{2} = - 6 a^{2} b^{2}

4 i^{3} a b^{3} = - 4 ia b^{3}

i^{4} b^{4} = 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

(5 x - x^{3})^{2} 8 + x^{2}

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

\frac{4}{( 2 y + 1 ) ^{2}}

6 a^{\frac{7}{4}} \cdot (a^{\frac{7}{4}} + a^{\frac{3}{4}})

(3 + 5)^{2}