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)

11 and 12 Grade Math 

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