Root of a complex number can be expressed in the standard form A + iB, where A and B are real.

In words we can say that any root of a complex number is a complex number

Let, z = x + iy be a complex number (x ≠ 0, y ≠ 0 are real) and n a positive integer. If the nth root of z be a then,

\(\sqrt[n]{z}\) = a

⇒ \(\sqrt[n]{x + iy}\) = a

⇒ x + iy = a\(^{n}\)

From the above equation we can clearly understand that

(i) a\(^{n}\) is real when a is purely real quantity and

(ii) a\(^{n}\) is either purely real or purely imaginary quantity when a is purely imaginary quantity.

We already assumed that, x ≠ 0 and y ≠ 0.

Therefore, equation x + iy = a\(^{n}\) is satisfied if and only if a is an imaginary number of the form A + iB where A ≠ 0and B ≠ 0 are real.

Therefore, any root of a complex number is a complex number.

Solved examples on roots of a complex number:

**1.** Find the square roots of -15 - 8i.

**Solution:**

Let \(\sqrt{-15 - 8i}\) = x + iy. Then,

\(\sqrt{-15 - 8i}\) = x + iy

⇒ -15 – 8i = (x + iy)\(^{2}\)

⇒ -15 – 8i = (x\(^{2}\) - y\(^{2}\)) + 2ixy

⇒ -15 = x\(^{2}\) - y\(^{2}\) .................. (i)

and 2xy = -8 .................. (ii)

Now (x\(^{2}\) + y\(^{2}\))\(^{2}\) = (x\(^{2}\) - y\(^{2}\))\(^{2}\) + 4x\(^{2}\)y\(^{2}\)

⇒ (x\(^{2}\) + y\(^{2}\))\(^{2}\) = (-15)\(^{2}\) + 64 = 289

⇒ x\(^{2}\) + y\(^{2}\) = 17 ................... (iii) [x\(^{2}\) + y\(^{2}\) > 0]

On Solving (i) and (iii), we get

x\(^{2}\) = 1 and y\(^{2}\) = 16

⇒ x = ± 1 and y = ± 4.

From (ii), 2xy is negative. So, x and y are of opposite signs.

Therefore, x = 1 and y = -4 or, x = -1 and y = 4.

Hence, \(\sqrt{-15 - 8i}\) = ± (1 - 4i).

**2.** Find the square root of i.

**Solution:**

Let √i = x + iy. Then,

√i = x + iy

⇒ i = (x + iy)\(^{2}\)

⇒ (x\(^{2}\) - y\(^{2}\)) + 2ixy = 0 + i

⇒ x\(^{2}\) - y\(^{2}\) = 0 .......................... (i)

And 2xy = 1 ................................. (ii)

Now, (x\(^{2}\) + y\(^{2}\))\(^{2}\) = (x\(^{2}\) - y\(^{2}\))\(^{2}\) + 4x\(^{2}\)y\(^{2}\)

(x\(^{2}\) + y\(^{2}\))\(^{2}\) = 0 + 1 = 1 ⇒ x\(^{2}\) + y\(^{2}\) = 1 ............................. (iii), [Since, x\(^{2}\) + y\(^{2}\) > 0]

Solving (i) and (iii), we get

x\(^{2}\) = ½ and y\(^{2}\) = ½

⇒ x = ±\(\frac{1}{√2}\) and y = ±\(\frac{1}{√2}\)

From (ii), we find that 2xy is positive. So, x and y are of same sign.

Therefore, x = \(\frac{1}{√2}\) and y = \(\frac{1}{√2}\) or, x = -\(\frac{1}{√2}\) and y = -\(\frac{1}{√2}\)

Hence, √i = ±(\(\frac{1}{√2}\) + \(\frac{1}{√2}\)i) = ±\(\frac{1}{√2}\)(1 + i)

**11 and 12 Grade Math****From Root of a Complex Number** **to HOME PAGE**

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.