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If $p=a+ib$, $q=c+id$ are points of the open disk $D(0,R)$, $R>0$, prove that at least one of the points $a+id, c+ib$ belong to the disk $D(0,R)$.

That is a problem of complex analysis in my university. The answer must be quite simple but i can't figure it out. Also it's my first time posting here so sorry if I did something wrong.

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    Do you mean that one of $a+ id$ and $c+ib$ must belong to $D(0,R)$? Try drawing a picture. Notice that $a+ib$, $c+ib$, $c+id$ and $a+id$ form a rectangle. – 2017-02-24
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    Also the post looks fine but don't forget to surround your maths with \$ signs, so that it renders properly. – 2017-02-24
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    I think that It means that one of them must be inside the circle.By that i mean that |p| or |q| must be – 2017-02-24

2 Answers 2

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Hint (assuming $a,b,c,d \in \mathbb{R}$): $\;|a+id|^2+|c+ib|^2=|p|^2+|q|^2 \lt 2R^2\,$.

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The disk $D(0, R)$ contains all points with magnitude less than $R$.

Since $p$ and $q$ are in $D(0,R)$, $|p|^2 = a^2 + b^2 < R^2$, and $|q|^2 = c^2 + d^2 < R^2$

Assume the point $a + id$ is not in the disk, which translates to $a^2 + d^2 \geq R^2$. Now play with these inequalities to show that the point $c + ib$ is in the disk.