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Popular >

simplify (1-3i)^4

Solution

simplify

(1 - 3 i)^{4}

Solution

28 + 96 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} : - 12 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (- 3 i)^{2} : 54 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (- 3 i)^{3} : - 108 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (- 3 i)^{4} : 81 i^{4}

= 1 - 12 i + 54 i^{2} - 108 i^{3} + 81 i^{4}

Simplify

1 - 12 i + 54 i^{2} - 108 i^{3} + 81 i^{4} : 96 i + 28

= 96 i + 28

Rewrite in standard complex form:

28 + 96 i

= 28 + 96 i

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Frequently Asked Questions (FAQ)

What is (1-3i)^4 ?
The solution to (1-3i)^4 is 28+96i