What is (1+isqrt(3))^7 ?

The solution to (1+isqrt(3))^7 is 64+64sqrt(3)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (1+isqrt(3))^7

simplify

(1 + i 3)^{7}

64 + 643 i

Solution steps

(1 + i 3)^{7}

Apply binomial theorem:

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

a = 1, b = i 3

= i = 0 \sum 7 (i 7) \cdot 1^{(7 - i)} (i 3)^{i}

Expand summation

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

Simplify

\frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 1^{7} (i 3)^{0} : 1

Simplify

\frac{7 !}{1 ! ( 7 - 1 ) !} \cdot 1^{6} (i 3)^{1} : 73 i

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 1^{5} (i 3)^{2} : 63 i^{2}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 1^{4} (i 3)^{3} : 1053 i^{3}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 1^{3} (i 3)^{4} : 315 i^{4}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 1^{2} (i 3)^{5} : 1893 i^{5}

Simplify

\frac{7 !}{6 ! ( 7 - 6 ) !} \cdot 1^{1} (i 3)^{6} : 189 i^{6}

Simplify

\frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 1^{0} (i 3)^{7} : 273 i^{7}

= 1 + 73 i + 63 i^{2} + 1053 i^{3} + 315 i^{4} + 1893 i^{5} + 189 i^{6} + 273 i^{7}

Simplify

1 + 73 i + 63 i^{2} + 1053 i^{3} + 315 i^{4} + 1893 i^{5} + 189 i^{6} + 273 i^{7} : 643 i + 64

= 643 i + 64

Rewrite in standard complex form:

64 + 643 i

= 64 + 643 i

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

\frac{1}{( s + 1 ) ( s + 10 )}

(\frac{w L}{2} - w a) a

\frac{p - 200}{3}

cos (\frac{( 2 n - 1 ) π x}{2})