changing a complex number into $a+bi$ form

Question

I saw a puzzle online today that read the following:

"Write the complex number $\frac{z+2i}{iz+3i}$ in the form $a+bi$"

I have tried multiplying by the complex conjugate and exponentialising the fraction but neither seem to help in putting it into cartesian form.

EDIT 1: Thank you for all the replies. Yes I mean for $z$ to be a complex variable.

EDIT 2: I have tried to reply to a user called @complexmanifold who tried to help initially. I told him that I did not know how to get from his second to his third line of working and I got voted down. I am sorry I am just trying to learn.

Answer 1

You know when you divide two complex numbers such as $\dfrac {2+3i}{4+5i}$, the standard thing is to just multiply by $\dfrac {4-5i}{4-5i}$ and this is guaranteed to work, because it is guaranteed to give you a real number in the denominator. Then it's just a matter of computing the multiplication on the top.

This problem is no different. Write $z$ as $x+yi$ (a technicality: I'm trying to avoid using $a, b$ since the directions say to write it in the form of $a+bi$, so the $a$ and $b$ should be the real and imaginary results of the answer, not of $z$).

So back to the topic; replace $z$ with $x+yi$ and then collect terms:

$$\dfrac{(x+yi)+2i}{i(x+yi)+3i} = \dfrac{x+(y+2)i}{-y+(x+3)i}.$$

And now it's a division problem just as in the beginning of the answer. So you can apply the same method.

And yeah this is the way to do it whenever you've got a division of two complex numbers, no matter if they are written in terms of $z$ like your original problem.

Answer 2

HINT:

\begin{align} \frac{z+2i}{iz+3i} &= \frac{-i(z+2i)}{z+3} \\ &= \frac{2-iz}{z+3} \\ &= \frac{[((a+3)-bi)(2-i(a+bi))]}{(a+3)^2+b^2} \\ &\vdots \end{align}

Answer 3

The real and imaginary parts of the solution are given by $(2a+3b+6)/(a^2+b^2+6a+9)$ and $(a^2+b^2+3a+2b)/(a^2+b^2+6a+9)$ respectively. Evidence of my research and workings done by others is shown in the comments on this page. Thanks to users complexmanifold and Peter Foreman for helping greatly to solve this problem.