Question

For complex values of z, solve Izl + z = (2+i).​

Answer 1

Answer:

For complex values of z, solve Izl + z = (2+i).

Answer 2

Step-by-step explanation:

Solution :

Let z = (x + iy).

Then,

$\: \: \: \: \: |z| + z = (2 + i)$

$➡ |x + iy| + (x + iy) = (2 + i)$

$➡(\sqrt{ {x}^{2} + {y}^{2} } + x) + iy = (2 + i)$

$➡ \sqrt{ {x}^{2} + {y}^{2} } + x = 2 \ \: and \: y = 1$

$➡ \sqrt{ {x}^{2} + {y}^{2} } = (2 - x) \: and \: y = 1$

$➡ \sqrt{ {x}^{2} + 1 } = (2 - x) \: and \: y = 1$

$➡( {x}^{2} + 1) = (2 - x) ^{2} \: and \: y = 1$

$➡( {x}^{2} + 1) = (4 + {x}^{2} - 4x) \: and \: y = 1$

$➡4 - 4x = 1 \: \: and \: \: y = 1$

$➡x = \frac{3}{4} \: \: and \: \: y = 1.$

Hence :

$z = ( \frac{3}{4} + i).$