UPSC NDA Math Model Paper 8

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)