+0  
 
+1
30
3
avatar+24 

Simplify (1 + i\sqrt{3})^6 (the i is outside of the sqrt of 3)

 Apr 30, 2022
 #1
avatar
0

Exponential form: \(1 + i \sqrt{3} = \sqrt{3} e^{\pi i/3}\)

 

\((\sqrt{3} e^{\pi i/3})^6 = 27 e^{2 \pi i} = \boxed{27}\)

 Apr 30, 2022
 #2
avatar+24 
+1

it's wrong, thank you tho 🫶

Celestianee  Apr 30, 2022
edited by Celestianee  Apr 30, 2022
 #3
avatar+1351 
+3

We have: \((1+ i \sqrt3)^6\)

 

Using the binomial theorem, we have: \(\large{{\sum\limits_{k=0}^n} {n \choose k} {a^{n-k}} b^k}\), where \(a=1\)\(b = i \sqrt3\), and \(n = 6\) 

 

We have to evaluate all cases from when \(0 \le k \le 6\):

 

\(k = 0: {6 \choose 0} \times 1^6 \times{ i \sqrt3} ^0 = 1\)

\(k = 1: {6 \choose 1} \times 1^5 \times{ i \sqrt3} ^1 = 6i \sqrt 3 \)

\(k = 2: {6 \choose 2} \times 1^4 \times{ i \sqrt3} ^2 = -45\)

\(k = 3: {6 \choose 3} \times 1^3 \times{ i \sqrt3} ^3 = -60i \sqrt 3 \)

\(k = 4: {6 \choose 4} \times 1^2 \times{ i \sqrt3} ^4 = 135\)

\(k = 5: {6 \choose 5} \times 1^1 \times{ i \sqrt3} ^5 = 54i \sqrt 3 \)

\(k = 6: {6 \choose 6} \times 1^0 \times{ i \sqrt3} ^6 = -27\)

 

To find the answer, you must add all these numbers.

 

Hope this helps!!! 

 Apr 30, 2022

29 Online Users

avatar