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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)