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Popular >

simplify (2-4i)^8

Solution

simplify

(2 - 4 i)^{8}

Solution

- 134912 - 86016 i

Solution steps

(2 - 4 i)^{8}

Apply binomial theorem:

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

a = 2, b = - 4 i

= i = 0 \sum 8 (i 8) \cdot 2^{(8 - i)} (- 4 i)^{i}

Expand summation

= \frac{8 !}{0 ! ( 8 - 0 ) !} \cdot 2^{8} (- 4 i)^{0} + \frac{8 !}{1 ! ( 8 - 1 ) !} \cdot 2^{7} (- 4 i)^{1} + \frac{8 !}{2 ! ( 8 - 2 ) !} \cdot 2^{6} (- 4 i)^{2} + \frac{8 !}{3 ! ( 8 - 3 ) !} \cdot 2^{5} (- 4 i)^{3} + \frac{8 !}{4 ! ( 8 - 4 ) !} \cdot 2^{4} (- 4 i)^{4} + \frac{8 !}{5 ! ( 8 - 5 ) !} \cdot 2^{3} (- 4 i)^{5} + \frac{8 !}{6 ! ( 8 - 6 ) !} \cdot 2^{2} (- 4 i)^{6} + \frac{8 !}{7 ! ( 8 - 7 ) !} \cdot 2^{1} (- 4 i)^{7} + \frac{8 !}{8 ! ( 8 - 8 ) !} \cdot 2^{0} (- 4 i)^{8}

Simplify

\frac{8 !}{0 ! ( 8 - 0 ) !} \cdot 2^{8} (- 4 i)^{0} : 256

Simplify

\frac{8 !}{1 ! ( 8 - 1 ) !} \cdot 2^{7} (- 4 i)^{1} : - 4096 i

Simplify

\frac{8 !}{2 ! ( 8 - 2 ) !} \cdot 2^{6} (- 4 i)^{2} : 28672 i^{2}

Simplify

\frac{8 !}{3 ! ( 8 - 3 ) !} \cdot 2^{5} (- 4 i)^{3} : - 114688 i^{3}

Simplify

\frac{8 !}{4 ! ( 8 - 4 ) !} \cdot 2^{4} (- 4 i)^{4} : 286720 i^{4}

Simplify

\frac{8 !}{5 ! ( 8 - 5 ) !} \cdot 2^{3} (- 4 i)^{5} : - 458752 i^{5}

Simplify

\frac{8 !}{6 ! ( 8 - 6 ) !} \cdot 2^{2} (- 4 i)^{6} : 458752 i^{6}

Simplify

\frac{8 !}{7 ! ( 8 - 7 ) !} \cdot 2^{1} (- 4 i)^{7} : - 262144 i^{7}

Simplify

\frac{8 !}{8 ! ( 8 - 8 ) !} \cdot 2^{0} (- 4 i)^{8} : 65536 i^{8}

= 256 - 4096 i + 28672 i^{2} - 114688 i^{3} + 286720 i^{4} - 458752 i^{5} + 458752 i^{6} - 262144 i^{7} + 65536 i^{8}

Simplify

256 - 4096 i + 28672 i^{2} - 114688 i^{3} + 286720 i^{4} - 458752 i^{5} + 458752 i^{6} - 262144 i^{7} + 65536 i^{8} : - 86016 i - 134912

= - 86016 i - 134912

Rewrite in standard complex form:

- 134912 - 86016 i

= - 134912 - 86016 i

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Frequently Asked Questions (FAQ)

What is (2-4i)^8 ?
The solution to (2-4i)^8 is -134912-86016i