Complex number question

[Acciara]

This is the question:
Give that z = 2-i / 1+i - 6+8i / u+i, find the values of u, u is a real number, such that Re z = Im z

Been on this question forever, I'll really appreciate if you explain the process of solving it to me.

p.s.

this is where I got to in the question, don't know how to continue:
2-i/1+i * 1-i/1-i = 1-3i / 2
6+8i / u+i * u-i / u-i = 6u + 8 + (8u-6)i / u^2 +1

 

[Deleted member 4993]

This is the question:
Give that z = 2-i / 1+i - 6+8i / u+i, find the values of u, u is a real number, such that Re z = Im z

Been on this question forever, I'll really appreciate if you explain the process of solving it to me.

As posted, your problem is:

\(\displaystyle \displaystyle \ z \ = \ 2 - \frac{i}{1} + i - 6 + \frac{8i}{u} + i \)

However, if your problem is:

\(\displaystyle \displaystyle z \ = \ \frac {2 - i}{1 + i} - \frac{6 + 8i}{u + i} \)

then you should have written it as:

z = (2-i) / (1+i) - (6+8i) / (u+i)

Tell us which of these are actually your problem and,

Please share your work with us .

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

 

[DrPhil]

This is the question:
Give that z = (2 - i)/(1 + i) - (6 + 8i)/(u + i), ....use parentheses to identify numerators and denominators
find the values of u, u is a real number, such that Re z = Im z

Been on this question forever, I'll really appreciate if you explain the process of solving it to me.

p.s.

this is where I got to in the question, don't know how to continue:
2-i/1+i * 1-i/1-i = (1 - 3i) / 2
6+8i / u+i * u-i / u-i = 6u + 8 + (8u-6)i / u^2 +1

Combining your two terms,

\(\displaystyle z = \dfrac{1-3i}{2} - \dfrac{(6u+8) + (8u +6)i}{u^2+1}\)

Separating real and imaginary parts,

\(\displaystyle z = \left[\dfrac{1}{2} - \dfrac{6u+8}{u^2+1}\right] + i\left[\dfrac{-3}{2} - \dfrac{8u +6}{u^2+1}\right]\)

Equate the real and imaginary parts, and solve the resulting quadratic equation for \(\displaystyle u\).

 

[Acciara]

Sorry I didn't use brackets, my bad.

But thank you Dr. Phil, that makes sense now.