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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