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Which complex number is a 6th root of i? A - HSC - SSCE Mathematics Extension 2 - Question 6 - 2018 - Paper 1

Question 6

Which complex number is a 6th root of i? A. − 1 √ 2 + 1 √ 2 i B. 1 √ 2 1 √ 2 i C. − √ 2 + √ 2 i D. − √ 2 − √ 2 i

Worked Solution & Example Answer:Which complex number is a 6th root of i? A - HSC - SSCE Mathematics Extension 2 - Question 6 - 2018 - Paper 1

Step 1

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Answer

To find the sixth roots of a complex number, we start by expressing the number in polar form. The complex number i can be represented as:

$i = e^{i rac{ heta}{2}}$ where $heta = \frac{\pi}{2} + 2k\pi$ , for $k = 0, 1, 2, 3, 4, 5$ (since we want 6 roots). This gives us:

$z_k = e^{i \left( \frac{\pi/2 + 2k\pi}{6} \right)} = e^{i \left( \frac{\pi}{12} + \frac{k\pi}{3} \right)}$

for $k = 0, 1, 2, 3, 4, 5$ .

Step 2

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Answer

Calculating for each k:

For $k = 0$ : $z_0 = e^{i \frac{\pi}{12}} = \cos\left(\frac{\pi}{12}\right) + i \sin\left(\frac{\pi}{12}\right)$
For $k = 1$ : $z_1 = e^{i \left( \frac{\pi}{12} + \frac{\pi}{3} \right)} = e^{i \frac{5\pi}{12}}$
For $k = 2$ : $z_2 = e^{i \left( \frac{\pi}{12} + \frac{2\pi}{3} \right)} = e^{i \frac{9\pi}{12}} = e^{i \frac{3\pi}{4}}$ (which corresponds to $rac{-\sqrt{2}}{2} + \frac{-\sqrt{2}}{2} i$ ) and so forth up to k=5.

Step 3

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Answer

After evaluating, we find:

A. \ $\left(-\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}} i\right)$ B. \ $\left(-\frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} i\right)$

C. \ $\left(-\sqrt{2} + \sqrt{2} i\right)$ D. \ $\left(-\sqrt{2} - \sqrt{2} i\right)$

Thus, the correct answer that matches our evaluated sixth roots of i is option A.