# Roots of a Complex Number

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)