# Complex number question

• Mar 25th 2010, 11:10 AM
Fandango1992
Complex number question
Hey,

I'm new as you can probably tell, so hello everyone! I'm doing AS Further Mathematics as an extra AS level during my second year at college, and I am finding it difficult to answer the following question in my homework:

Find the complex number $\displaystyle z$ which satisfies $\displaystyle (2+i)z+(3-2i)z^*=32$ .

How would I go about attempting this? I'm unsure of what step/s to take first.

Thanks very much for any help! (Happy)
• Mar 25th 2010, 11:51 AM
running-gag
Hi

Quote:

Originally Posted by Fandango1992
Find the complex number $\displaystyle z$ which satisfies $\displaystyle (2+i)z+(3-2i)z^*$

The question is not complete : which satisfies what ?
• Mar 26th 2010, 01:37 AM
Fandango1992
Quote:

Originally Posted by running-gag
Hi

The question is not complete : which satisfies what ?

Oops, sorry! I have recorrected it. Must've been shattered...
• Mar 26th 2010, 04:11 AM
HallsofIvy
z* is the complex conjugate of z?

Let z= x+ iy. Then your equation is (2+i)(x+ iy)+(3-2i)(x- iy)=32. Multiplying that out, 2x- y+ i(x+ 2y)+ 3x- 2y+ i(-2x- 3y)= 32. Set the real part of the left side equal to 32 and the imaginary part equal to 0. You will have two equations to solve for x and y.
• Mar 26th 2010, 12:40 PM
running-gag
Just for fun you can solve it another way
$\displaystyle (2+i)z+(3-2i) \bar{z}=32$

Take the conjugate of both sides
$\displaystyle (2-i)\bar{z}+(3+2i) z=32$

This leads to the following system
$\displaystyle (2+i)z+(3-2i) \bar{z}=32$
$\displaystyle (3+2i) z + (2-i)\bar{z}=32$
that you can solve for z by eliminating $\displaystyle \bar{z}$
• Mar 26th 2010, 04:45 PM
Fandango1992
Thanks a lot for your help! Now I can move to question two, haha. Good times...