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Popular >

simplify (6-2i)^4

Solution

simplify

(6 - 2 i)^{4}

Solution

448 - 1536 i

Solution steps

(6 - 2 i)^{4}

Apply binomial theorem:

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

a = 6, b = - 2 i

= i = 0 \sum 4 (i 4) \cdot 6^{(4 - i)} (- 2 i)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 6^{4} (- 2 i)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 6^{3} (- 2 i)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 6^{2} (- 2 i)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 6^{1} (- 2 i)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 6^{0} (- 2 i)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 6^{4} (- 2 i)^{0} : 1296

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 6^{3} (- 2 i)^{1} : - 1728 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 6^{2} (- 2 i)^{2} : 864 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 6^{1} (- 2 i)^{3} : - 192 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 6^{0} (- 2 i)^{4} : 16 i^{4}

= 1296 - 1728 i + 864 i^{2} - 192 i^{3} + 16 i^{4}

Simplify

1296 - 1728 i + 864 i^{2} - 192 i^{3} + 16 i^{4} : - 1536 i + 448

= - 1536 i + 448

Rewrite in standard complex form:

448 - 1536 i

= 448 - 1536 i

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Frequently Asked Questions (FAQ)

What is (6-2i)^4 ?
The solution to (6-2i)^4 is 448-1536i