Understanding Complex Numbers in Equations

[lamefeed]

Homework Statement

So the problem I have is this silly little equation..
$$\frac {z - 7}{z + 3} = i $$

Homework Equations

This is the thing, I don't think you need anything more advanced than basic algebra to solve this problem.

The Attempt at a Solution

And I've tried solving it doing the following:

$$\frac {z - 7}{z + 3} = i $$
Adding $z+3$ to both sides ending up with:
$$z - 7 = i(z + 3) $$
Solving the parenthesis
$$z - 7 = iz + 3i $$
Subtracting the 7
$$z = iz + 3i + 7$$This is where I'm getting stuck, would anyone be kind and explain where I've done something wrong?
According to the Answers sheet I'm supposed to end up with $z = 2 + 5i$. I don't really care about the correct answer what I want to know is how to get there without screwing up.

Cheers!

 

[member 587159]

I didn't check any of your calculations, but proceeding from where you are, you have:

$z = iz + 3i + 7 \iff z - iz = 3i + 7 \iff z(1-i) = 3i + 7 \iff z = \frac{3i+7}{1-i}$

And this can be simplified by multiplying nominator and denominator with $1+i$ (details left for you), resulting in $z = 2 + 5i$, as desired.

 

[fresh_42]

You have to be more careful with division. For complex numbers we have $|z|^2=z\cdot \bar{z}$ which means $\dfrac{1}{z}=z^{-1}=\dfrac{\bar{z}}{|z|^2}$.

If you go ahead you with what you have and which is correct, you get $z=\dfrac{7+3i}{1-i}$. Do you know where to go from there?

Here is an insight article about complexx numbers, which I think you should read:
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

 

[member 587159]

You have to be more careful with division. For complex numbers we have $|z|^2=z\cdot \bar{z}$ which means $\dfrac{1}{z}=z^{-1}=\dfrac{\bar{z}}{|z|^2}$.

Here is an insight article about complexx numbers, which I think you should read:
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

Where does the OP perform division?

 

[m4r35n357]

As far as I can tell you are on the right track - you just haven't finished yet!

[EDIT] OK. so post #2 has let the cat out of the bag!

 

[fresh_42]

Where does the OP perform division?

$(z+3)^{-1}$ is a division. And latest at the final quotient, he will need this formula.

 

[member 587159]

$(z+3)^{-1}$ is a division. And latest at the final quotient, he will need this formula.

I don't see the problem. As long as $z \neq -3$, OP should be fine.

 

[fresh_42]

I don't see the problem. As long as $z \neq -3$, OP should be fine.

Yes, he was correct, as edited in above. I just wanted to say that division in the complex numbers takes more care than usual. And one division has to be made, no matter where, but at least one is needed.

 

[lamefeed]

I've solved it, I'll take a look at the article fresh_42.

The steps I took where the following:
$$z=\frac{(3i+7)(1+i)}{(1-i)(1+i)} \Rightarrow \frac{3i+3i^2+7+7i}{1+i-i-i^2} \Rightarrow \frac{4+10i}{2} \Rightarrow 2+5i$$

Been struggling with that one for some time... Not sure why don't feel like it was particularly hard. Thanks to you guys I'm able to get further into my math homework.

Cheers!

 

[SammyS]

Homework Statement

So the problem I have is this silly little equation..
$$\frac {z - 7}{z + 3} = i $$

Another way to do this is to simplify $\displaystyle \ \frac {z - 7}{z + 3} \$ by a method often used by @Mark44 , one of our Staff Mentors .

The idea is that $\displaystyle \ \frac {z - 7}{z + 3} \$ can be expressed in a way so that z appears only once, in the denominator, rather than in both numerator and denominator. That makes it easy to solve for $z$ .

$\displaystyle \frac {z - 7}{z + 3}=\frac {z+3-3 - 7}{z + 3} \$

$\displaystyle =\frac {z +3}{z + 3} + \frac {-10}{z + 3}$

$\displaystyle =1- \frac {10}{z + 3}$​

So your equation becomes $\displaystyle \ \ \ 1- \frac {10}{z + 3}=i \,.\$

Resist any temptation to multiply through by $(z+3)$ . That would just defeat the purpose of doing the above work. Keep $(z+3)$ together until the next to last step.

 

[m4r35n357]

Great, now I can't solve it any more!

 

[Merlin3189]

Now you've solved it, may I just ask for clarification of this step please?

... proceeding from where you are, you have:
$z = iz + 3i + 7 \iff - iz = 3i + 7$.

I can't see what operation you performed here.
It appears to me that you should get $z = iz + 3i + 7 \iff z - iz = 3i + 7$
and your line suggests to me that z=0 ?

 

[member 587159]

Now you've solved it, may I just ask for clarification of this step please?

I can't see what operation you performed here.
It appears to me that you should get $z = iz + 3i + 7 \iff z - iz = 3i + 7$
and your line suggests to me that z=0 ?

Typo, I accidentally left out the z left. Your correction is right.

I corrected that post. Thanks.

 

[SammyS]

Great, now I can't solve it any more!

@m4r35n357 What post was this referencing ?

 

[lamefeed]

Thanks for all the good suggestions :)!
Starting to like doing math with complex numbers :)

 

[SammyS]

Thanks for all the good suggestions :)!
Starting to like doing math with complex numbers, been struggling with another one though, not sure if I should make a new thread or keep it within this one?

...

Cheers!

Yes, start a new thread.

 

[fresh_42]

Thanks for all the good suggestions :)!
Starting to like doing math with complex numbers :)

sign error

 

[StoneTemplePython]

another approach is write out $z = a +bi$

1.) recall that magnitudes (like determinants) multiply. So take the squared magnitude of each side of the original equation and get

$\big((a-7)^2 + b^2\big) \big((a+3)^2 + b^2\big)^{-1} = 1 \to (a-7)^2 = (a+3)^2$

which tells us
$(a-7) = -(a+3) \to a = 2$

(note it cannot be $(a-7) = +(a+3)$ because that is equivalent to saying $-7 = +3$) 2.) plugging back into the original problem

$\big(-5 +bi\big) \big(5 + bi\big)^{-1} = i$

you can directly chug through the inversion here, but there's a nicer finish. Multiply each side by $-i$ (that way, symbolically this reads $x \cdot x^{-1} = 1$ which is a typical way for relating something to its inverse)

$\big(b + 5i \big) \big(5 + bi\big)^{-1} = 1$

By eyeballing this, we can guess $b= 5$ and see that works. Since each non-zero complex number has a unique inverse, then the guess is in fact the solution.

 

[m4r35n357]

@m4r35n357 What post was this referencing ?

Oops, #10. I don't get it!

 

[SammyS]

Oops, #10. I don't get it!

In Post #10, the last step I showed gave :

So your equation becomes $\displaystyle \ \ \ 1- \frac {10}{z + 3}=i \,.\$​

From there, isolate the.term with z+3 .

$\displaystyle 1-i= \frac {10}{z + 3} \,\$​

.then

$\displaystyle z + 3= \frac {10}{1-i} \,\$​

Rationalize the denominator & simplify.

 

[m4r35n357]

$\displaystyle z + 3= \frac {10}{1-i} \,\$​

Rationalize the denominator & simplify.

Ah, OK thanks. I would have counted the "exchange" of terms as equivalent to (potentially) multiplying by 0, just like the solution I got, but then I'm no mathematician!