Let √8−6i=x+iy Squaring both sides, we get ⇒8−6i=(x+iy)2 ⇒8−6i=x2+(iy)2+i2xy ⇒8−6i=x2−y2+i2xy Compare real and imaginary parts, ∴x2−y2=8 and 2xy=−6 2xy=−6⇒y=−3∕x .......(1) Now, x2−y2=8 ⇒x2−(9∕x2)=8 ⇒x4−9=8x2 ⇒x4−8x2−9=0 ⇒x4−9x2+x2−9=0 ⇒x2(x2−9)+1(x2−9)=0 ⇒(x2+1)(x2−9)=0 So, (x2+1)=0⇒x2≠−1 And (x2−9)=0 ∴x=±3 Put the value of x in equation 1st If x=3 than y=−1 If x=−3 than y=1 So, square root of (8−6i)=(3−i) or (−3+i) ∴√8−6i=±(3−i)