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# help <3

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

Apr 30, 2022

#1
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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
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it's wrong, thank you tho 🫶

Celestianee  Apr 30, 2022
edited by Celestianee  Apr 30, 2022
#3
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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