Question

find the solution of z^2(1-z^2)=16 ,where z is a complex number??

Answer 1

Answer:

Step-by-step explanation:

Let z = x + iy. Then,

z2=z¯⇒(x+iy)2=x−iy

⇒ (x2−y2)+2ixy=x−iy. ...(i)

Equating real parts and imaginary parts on both sides of (i) separately, we get

x2−y2=x ...(ii)and2xy=−y ...(iii).

From (iii), we get

2xy+y=0⇒y(2x+1)=0⇒y=0orx=−12.

Case I When y = 0.

Putting y = 0 in (ii), we get

x2−x=0⇒x(x−1)=0⇒x=0orx=1.

∴ (x=0,y=0)or(x=1,y=0)

Thus, z=(0+i0)orz=(1+i0).

Case II When x=−12.

Putting x=−12 in (ii), we get

(−12)2−y2=(−12)⇒y2=(14+12)=34⇒y=±3–√2.

∴ (x=−12,y=3–√2)or(x=−12,y=−3–√2).

Thus, z=(−12+3–√2i)orz=(−12−3–√2i).

Hence, z=0,1(−12+3–√2i)and(−12−3–√2i) are the required roots of the given equation.