Question

for complex number z= 2+i/3+i, then [z] is (please also tell how did you do it)

Answer 1

Answer:

$z = \frac{2 + i}{3 + i} \\ z = \frac{2 + i}{3 + i} \times \frac{3 - i}{3 - i} \\ z = \frac{(2 + i)(3 - i)}{ {3}^{2} - i {}^{2} } \\ = \frac{6 + 3i - 2i - {i}^{2} }{9 - ( - 1)} \\ = \frac{6 + i - ( - 1)}{10} \\ = \frac{7 + i}{10} \\ z= \frac{7}{10} + \frac{i}{10}$

Answer 2

2+I/3+I rationalising by 3-i
=6-2i +3i+1/ 9+1
= 7/10+I/10