Let −15−8i−−−−−−−√ = x + iy. Then,
−15−8i−−−−−−−√ = x + iy
⇒ -15 – 8i = (x + iy)2
⇒ -15 – 8i = (x2 - y2) + 2ixy
⇒ -15 = x2 - y2 .................. (i)
and 2xy = -8 .................. (ii)
Now (x2 + y2)2 = (x2 - y2)2 + 4x2y2
⇒ (x2 + y2)2 = (-15)2 + 64 = 289
⇒ x2 + y2 = 17 ................... (iii) [x2 + y2 > 0]
On Solving (i) and (iii), we get
x2 = 1 and y2 = 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, −15−8i−−−−−−−√ = ± (1 - 4i).
2. Find the square root of i.
Solution:
Let √i = x + iy. Then,
√i = x + iy
⇒ i = (x + iy)2
⇒ (x2 - y2) + 2ixy = 0 + i
⇒ x2 - y2 = 0 .......................... (i)
And 2xy = 1 ................................. (ii)
Now, (x2 + y2)2 = (x2 - y2)2 + 4x2y2
(x2 + y2)2 = 0 + 1 = 1 ⇒ x2 + y2 = 1 ............................. (iii), [Since, x2 + y2 > 0]
Solving (i) and (iii), we get
x2 = ½ and y2 = ½
⇒ x = ±1√2 and y = ±1√2
From (ii), we find that 2xy is positive. So, x and y are of same sign.
Therefore, x = 1√2 and y = 1√2 or, x = -1√2 and y = -1√2
Hence, √i = ±(1√2 + 1√2i) = ±1√2(1 + i)
Written on March 11, 2018, 7:23 p.m.