What is (sqrt(2)-3i)^4 ?

The solution to (sqrt(2)-3i)^4 is -23+84sqrt(2)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (sqrt(2)-3i)^4

simplify

(2 - 3 i)^{4}

- 23 + 842 i

Solution steps

(2 - 3 i)^{4}

Apply binomial theorem:

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

a = 2, b = - 3 i

= i = 0 \sum 4 (i 4) (2)^{(4 - i)} (- 3 i)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (2)^{4} (- 3 i)^{0} : 4

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (2)^{3} (- 3 i)^{1} : - 242 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (2)^{2} (- 3 i)^{2} : 108 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (2)^{1} (- 3 i)^{3} : - 1082 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (2)^{0} (- 3 i)^{4} : 81 i^{4}

= 4 - 242 i + 108 i^{2} - 1082 i^{3} + 81 i^{4}

Simplify

4 - 242 i + 108 i^{2} - 1082 i^{3} + 81 i^{4} : 842 i - 23

= 842 i - 23

Rewrite in standard complex form:

- 23 + 842 i

= - 23 + 842 i

(- sin (x) u)^{2}

(2 k - 3) (3 k - 3)^{'}

(x - 3 + 2 i) (x - 3 - 2 i)

\frac{2 x + 2}{( x - 1 ) ( x + 3 )}

(7 x - 1) (x^{2} - 2 x + 1)^{''}