Math, asked by ayushsingh39as, 7 months ago

Q.34. Solve the equation z^2=z', where z is a complex
number. (z'=z bar).

Answers

Answered by shreyasinha28
4

Answer:

The equation solution z^2 = zˉ ,where z is a complex number has 4 solutions.

Step-by-step explanation:

z2 = z ⇒ x2 – y2 + i2xy = x – iy

Therefore, x2 – y2 = x ...(1)

and 2xy = – y ... (2)

From (2), we have y = 0 or x = -1/2

When y = 0, from (1), we get

x2 – x = 0, i.e., x = 0 or x = 1.

When x =-1/2, from (1), we get

y2 = 1/4 +1/2 or y2 =3/4,

i.e., y =±√3/2

Answered by adityamulge257
3

Answer:

z^2 =z= x^2 -y square +12xy=x-y

We have y=0 or x=-1/2

when y =0, we get x^2-x=0 i.e x=0 or x=1

when x =-1/2,we get y square 1/4+1/2

solutions of given equation are

0+i0,1+i0 ,-1/2+i √3/2,-1/2-i √3/2

Step-by-step explanation:

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