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

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.

did you work it out?

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

Think I got something wrong :/

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

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

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

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

did it help?

Yes, thank you.

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