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Here is my approach: Suppose the parallelogram is in the first quadrant. Then, we can say that $z_1=x_1+iy_1$, $z_2=x_2+iy_2$, and $z_3=x_3+iy_3$. Let $z_1$ be in bottom left, $z_2$ be in top left, $z_3$ be in top right. We know that opposite sides of a parallelogram are parallel and congruent. Then, $z_4=[x_1+|x_3-x_2|]+iy_1$.
Now, to continue, I would consider a parallelogram that is totally contained in each quadrant, then start looking at the cases where the parallelogram is partly contained in quadrant..... this seems like a lot of cases. I was wondering if there would be an elementary way that I could generalize this problem down to a few cases, or even just a single case. Thank you.

User7238
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  • Um, more or less. I think from that I can see how this reduces down and we only consider the "one" case. As always, thank you, Martin! – User7238 Mar 02 '20 at 11:10
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    I was surprised that someone downvoted the OP's query. I thought that his query showed reasonable work but that he was stuck on cartesian rather than polar coordinates. Therefore, I upvoted his query back. – user2661923 Mar 02 '20 at 12:10
  • Thank you very much! – User7238 Mar 02 '20 at 12:11

2 Answers2

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I agree with and upvoted Martin R's comment and tomi's answer. My answer is intended to provide an alternate way of developing the OP's intuition, which may have been the point of the problem.

Every non-zero complex number can be expressed as $\;re^{i\theta} = r[\cos \,\theta + i\sin \,\theta],\;$ where $\,r\,$ is the magnitude of the number, and $\;\theta\;$ represents the angle between the non-zero complex number and the origin. For $\;z_1, z_2, z_3, z_4\;$ to form a parallelogram, with $\,z_4\,$ opposite of $\,z_2,\,$ two things must happen:

(1) The magnitude of $\;(z_4 - z_1)\;$ = the magnitude of $\;(z_3 - z_2).$
(2) The angle formed by the vector $\;(z_4 - z_1)\;$ = the angle formed by the vector $\;(z_3 - z_2).$

Therefore, if $\;(z_3 - z_2)\; = \;re^{i\theta},\;$ then $\;(z_4 - z_1)\;$ must also = $\;re^{i\theta}.$

user2661923
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You can think of the complex numbers as being position vectors.

The vector going from $z_2$ to $z_3$ is $z_3-z_2$.

The vector going from $z_2$ to $z_1$ is $z_1-z_2$.

Because it's a parallelogram, the vector going from $z_2$ to $z_4$ is the sum of those two vectors, which is $z_3-z_2+z_1-z_2=z_3+z_1-2z_2$.

The position vector of $z_4$ is given by $z_2+z_3+z_1-2z_2=z_3+z_1-z_2$.

tomi
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