What is (sqrt(3)-i)^4 ?

The solution to (sqrt(3)-i)^4 is -8-8sqrt(3)i

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simplify (sqrt(3)-i)^4

simplify

(3 - i)^{4}

- 8 - 83 i

Solution steps

(3 - i)^{4}

Apply binomial theorem:

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

a = 3, b = - i

= i = 0 \sum 4 (i 4) (3)^{(4 - i)} (- i)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (3)^{4} (- i)^{0} : 9

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (3)^{3} (- i)^{1} : - 123 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (3)^{2} (- i)^{2} : 18 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (3)^{1} (- i)^{3} : - 43 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (3)^{0} (- i)^{4} : i^{4}

= 9 - 123 i + 18 i^{2} - 43 i^{3} + i^{4}

Simplify

9 - 123 i + 18 i^{2} - 43 i^{3} + i^{4} : - 83 i - 8

= - 83 i - 8

Rewrite in standard complex form:

- 8 - 83 i

= - 8 - 83 i

(1 - 2 i)^{5}

(\frac{4}{16} x + \frac{5}{16} y + \frac{3}{16} z + \frac{4}{16} u)^{4}

\frac{- x ^{2} + y ^{2}}{( x ^{2} + y ^{2} ) ^{2}}

(2 + i)^{5}

\frac{( 1 + j ^{3} ) ( 1 + j ^{7} )}{( 7 - j ^{9} ) ( 2 - j )}