Complex numbers question
- Thread starter dhruv
- Start date
D
Deleted member 4993
Guest
Please follow the rules of posting at this forum, enunciated at:
https://www.freemathhelp.com/forum/threads/read-before-posting.109846/
Please share your work/thoughts and context of the problem (what is the subject topic?) - so that we know where to begin to help you.
Sorry for the inconvenience! I believe what I didn't include is what I've already tried.
I have attempted the problem in two ways:
1. I first rationalized the fraction as it is (with z as z), but I got no where even after substituting z for x + iy afterwards.
2. I first substituted z for x + iy and then tried to simplify it leading to just a very complicated fraction.
I have attempted the problem in two ways:
1. I first rationalized the fraction as it is (with z as z), but I got no where even after substituting z for x + iy afterwards.
2. I first substituted z for x + iy and then tried to simplify it leading to just a very complicated fraction.
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
Learn this important and most useful identity:
\(\displaystyle \large\frac{1}{z}=\frac{\overline{\,z\,}}{|z|^2}\)
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,081
What we want to see from you is actual work, not just a general description -- you may have tried doing the right thing, but made some little mistake, or you may not mean the right thing though the words are right.
What does it mean to "rationalize" a fraction? Normally we rationalize the denominator, when the denominator contains radicals. Do you mean something similar, like "realizing" the denominator (making it a real number using the complex conjugate)? Show us what you got. Then show us what you got after the substitution, so we can either correct it, or make a suggestion for simplifying it.
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
\(\displaystyle z+i=x+i(y+1)\\z+2=(x+2)\\\overline{z+2}=\overline{z}+2=(x+2)-iy\)
\(\displaystyle z\cdot\overline{z}=|z|^2=x^2+y^2\)
\(\displaystyle (z+i)(\overline{z}+2)=z\cdot\overline{z}+i\overline{z}+2z+2i=(x^2+y^2)+xi+y+2x+2yi+2i\)
\(\displaystyle \Re[(z+i)(\overline{z}+2)]=(x^2+y^2)+y+2x\)
Can you finish?
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,081
Looks good.
The work in the denominator can be done much more easily, if you keep it as [(x+2) + yi][(x+2) - yi]. The product of (a+b)(a-b) is a^2 - b^2, so you just get (x+2)^2 - (yi)^2 = (x+2)^2 + y^2, which is the form they asked for. You never have to expand fully. The numerator does need to be expanded as you did. (Of course, doing it for myself rather than to show someone, I would write out a lot less.)
The work in the denominator can be done much more easily, if you keep it as [(x+2) + yi][(x+2) - yi]. The product of (a+b)(a-b) is a^2 - b^2, so you just get (x+2)^2 - (yi)^2 = (x+2)^2 + y^2, which is the form they asked for. You never have to expand fully. The numerator does need to be expanded as you did. (Of course, doing it for myself rather than to show someone, I would write out a lot less.)
Thanks a lot for the help and putting in your valuable time!Check out the solution i've wrote at the following file.
Thanks, Dr. Peterson! Yes, that's much more easy... I didn't realize I could do that, tough rsrs.
I'm happy to help! By the way, I'm sorry for any grammar mistakes I may did. I'm brazilian and don't speak English fluently, so I probably made some of them...