Find the square roots of the following
(i) -31
(ii) -32
(iii) -36
(iv) -144
(v) `-(16)/(25)`
(vi) `-(8)/(729)`.

Answer

To find the square roots of the given negative numbers, we will use the concept of complex numbers. The square root of a negative number can be expressed in terms of the imaginary unit \(i\), where \(i = \sqrt{-1}\).

### Step-by-Step Solution:

**(i) Find the square root of -31:**

1. Write \(-31\) as \(-1 \times 31\).
2. Apply the square root:
\[ \sqrt{-31} = \sqrt{-1} \times \sqrt{31} = i\sqrt{31} \]
3. Therefore, the square roots of \(-31\) are:
\[ \pm i\sqrt{31} \]

**(ii) Find the square root of -32:**

1. Write \(-32\) as \(-1 \times 32\).
2. Apply the square root:
\[ \sqrt{-32} = \sqrt{-1} \times \sqrt{32} = i\sqrt{32} \]
3. Simplify \(\sqrt{32}\):
\[ \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \]
4. Therefore, the square roots of \(-32\) are:
\[ \pm 4i\sqrt{2} \]

**(iii) Find the square root of -36:**

1. Write \(-36\) as \(-1 \times 36\).
2. Apply the square root:
\[ \sqrt{-36} = \sqrt{-1} \times \sqrt{36} = i\sqrt{36} \]
3. Simplify \(\sqrt{36}\):
\[ \sqrt{36} = 6 \]
4. Therefore, the square roots of \(-36\) are:
\[ \pm 6i \]

**(iv) Find the square root of -144:**

1. Write \(-144\) as \(-1 \times 144\).
2. Apply the square root:
\[ \sqrt{-144} = \sqrt{-1} \times \sqrt{144} = i\sqrt{144} \]
3. Simplify \(\sqrt{144}\):
\[ \sqrt{144} = 12 \]
4. Therefore, the square roots of \(-144\) are:
\[ \pm 12i \]

**(v) Find the square root of \(-\frac{16}{25}\):**

1. Write \(-\frac{16}{25}\) as \(-1 \times \frac{16}{25}\).
2. Apply the square root:
\[ \sqrt{-\frac{16}{25}} = \sqrt{-1} \times \sqrt{\frac{16}{25}} = i\sqrt{\frac{16}{25}} \]
3. Simplify \(\sqrt{\frac{16}{25}}\):
\[ \sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} \]
4. Therefore, the square roots of \(-\frac{16}{25}\) are:
\[ \pm \frac{4}{5}i \]

**(vi) Find the square root of \(-\frac{8}{729}\):**

1. Write \(-\frac{8}{729}\) as \(-1 \times \frac{8}{729}\).
2. Apply the square root:
\[ \sqrt{-\frac{8}{729}} = \sqrt{-1} \times \sqrt{\frac{8}{729}} = i\sqrt{\frac{8}{729}} \]
3. Simplify \(\sqrt{\frac{8}{729}}\):
\[ \sqrt{\frac{8}{729}} = \frac{\sqrt{8}}{\sqrt{729}} = \frac{2\sqrt{2}}{27} \]
4. Therefore, the square roots of \(-\frac{8}{729}\) are:
\[ \pm \frac{2\sqrt{2}}{27}i \]

### Final Answers:
1. \(-31\): \(\pm i\sqrt{31}\)
2. \(-32\): \(\pm 4i\sqrt{2}\)
3. \(-36\): \(\pm 6i\)
4. \(-144\): \(\pm 12i\)
5. \(-\frac{16}{25}\): \(\pm \frac{4}{5}i\)
6. \(-\frac{8}{729}\): \(\pm \frac{2\sqrt{2}}{27}i\)