Question

Find the complex number z which satisfies (2+i)z+(3-2i)z*=32

Answer 1

so assume z = a+bi.
(2+i)(a+bi)+(3-2i)((a+bi) = 32
(2a - b) + (a + 2b)i + (3a + 2) + (-2a +3b)i = 32
and (2a - b) + (3a + 2) = 32 (the real number parts)
and (a + 2b) + (-2a + 3b) = 0 (the imaginary parts)

solve for a and b. you have a system with 2 equations and 2 unknowns.

Answer 2

did you work it out?

Answer 3

I'm currently double checking it through the equation. For z I have 80/13 + 16/13i =z

Answer 4

Think I got something wrong :/

Answer 5

so the first eq: 5a + b = 32
the second eq: -a + 5b = 0

so 11b = 32 and b = 32/11

-26a = -160 so a = 80/13

Answer 6

oops... 26b = 32 so b = 16/13

Answer 7

let's see...

\[(2+i)(5+i) +(3-2i)(5+i)= 2(13)\Rightarrow (10 - 1) + 7i + (15 +2) - 7i = 26\]
\[\Rightarrow 9 + 17 + 7i - 7i = 26 \surd\]

Answer 8

Thanks a lot. Had to bash my brains for an hour going back and forth, going nowhere to finally result to using this site :)

Answer 9

did it help?

Answer 10

Yes, thank you.

Answer 11

good. there are lot's of helpful people here so come anytime you have a question or need some insight.