What is (1-sqrt(3)i)^4 ?

The solution to (1-sqrt(3)i)^4 is -8+8sqrt(3)i

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simplify (1-sqrt(3)i)^4

simplify

(1 - 3 i)^{4}

- 8 + 83 i

Solution steps

(1 - 3 i)^{4}

Apply binomial theorem:

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

a = 1, b = - 3 i

= i = 0 \sum 4 (i 4) \cdot 1^{(4 - i)} (- 3 i)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1^{4} (- 3 i)^{0} : 1

Simplify

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

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (- 3 i)^{2} : 18 i^{2}

Simplify

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

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (- 3 i)^{4} : 9 i^{4}

= 1 - 43 i + 18 i^{2} - 123 i^{3} + 9 i^{4}

Simplify

1 - 43 i + 18 i^{2} - 123 i^{3} + 9 i^{4} : 83 i - 8

= 83 i - 8

Rewrite in standard complex form:

- 8 + 83 i

= - 8 + 83 i

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

(x - 2) (x - 5) (x - 3) (x + 3)

\frac{1}{( s ^{2} + a ^{2} ) ( s ^{2} + b ^{2} )}

(y + 2) (y - 2)

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