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

lo g_{10} (7 z)

5 (10 - 42)

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

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