Q. 28

Find the complex number z for which |z| = z + 1 + 2i.

Answer :

Given: |z| = z + 1 + 2i

Consider,


|z| = (z + 1) + 2i


Squaring both the sides, we get


|z|2 = [(z + 1) + (2i)]2


|z|2 = |z + 1|2 + 4i2 + 2(2i)(z + 1)


|z|2 = |z|2 + 1 + 2z + 4(-1) + 4i(z + 1)


0 = 1 + 2z – 4 + 4i(z + 1)


2z – 3 + 4i(z + 1) = 0


Let z = x + iy


2(x + iy) – 3 + 4i(x + iy + 1) = 0


2x + 2iy – 3 + 4ix + 4i2y + 4i = 0


2x + 2iy – 3 + 4ix + 4(-1)y + 4i = 0


2x – 3 – 4y + i(4x + 2y + 4) = 0


Comparing the real part, we get


2x – 3 – 4y = 0


2x – 4y = 3 …(i)


Comparing the imaginary part, we get


4x + 2y + 4 = 0


2x + y + 2 = 0


2x + y = -2 …(ii)


Subtracting eq. (ii) from (i), we get


2x – 4y – (2x + y) = 3 – (-2)


2x – 4y – 2x – y = 3 + 2


-5y = 5


y = -1


Putting the value of y = -1 in eq. (i), we get


2x – 4(-1) = 3


2x + 4 = 3


2x = 3 – 4


2x = - 1



Hence, the value of z = x + iy




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