Problem 14
Medium Algebra of Complex Numbers

If |a|-|b|=1 , show that there do not exist complex numbers…

If $|a|-|b|=1$, show that there do not exist complex numbers $a$ and $b$ for which $\text{Re}(a)=2\,\text{Im}(b)$ and $\text{Im}(a)=2\,\text{Re}(b)$.

Explanation of Method

Write $a=x_1+iy_1$, $b=x_2+iy_2$ and translate all three conditions into real equations. Substituting the two linear conditions into the modulus of $a$ gives a direct relationship between $|a|$ and $|b|$; combining that with the given difference condition either forces a contradiction, or pins down explicit values of $|a|$ and $|b|$ that can be checked directly for consistency.

Step 1 · Set up coordinates

Let $a = x_1+iy_1$, $b=x_2+iy_2$ with $x_1,y_1,x_2,y_2\in\mathbb R$. The three conditions become:

$$ \sqrt{x_1^2+y_1^2} - \sqrt{x_2^2+y_2^2} = 1, \qquad x_1 = 2y_2, \qquad y_1 = 2x_2 $$

Step 2 · Combine the linear conditions

Substituting $x_1=2y_2$ and $y_1=2x_2$ into $|a|^2=x_1^2+y_1^2$:

$$ |a|^2 = (2y_2)^2+(2x_2)^2 = 4(x_2^2+y_2^2) = 4|b|^2 \quad\Longrightarrow\quad |a| = 2|b| $$

Step 3 · Combine with the modulus condition

Substituting $|a|=2|b|$ into $|a|-|b|=1$:

$$ 2|b|-|b| = 1 \quad\Longrightarrow\quad |b|=1, \qquad |a| = 2 $$

This is not a contradiction — it's a perfectly consistent pair of values. So, as literally stated, the three conditions do not rule each other out algebraically; we should check directly whether a matching pair $(a,b)$ actually exists.

Step 4 · An explicit example

Take $b=1$ (so $x_2=1,\ y_2=0$, and indeed $|b|=1$). Then the linear conditions require $x_1=2y_2=0$ and $y_1=2x_2=2$, i.e. $a=2i$.

Checking all three conditions:

More generally, for any $\theta$, $b=\cos\theta+i\sin\theta$ (so $|b|=1$) and $a = 2\sin\theta+2i\cos\theta$ (so $|a|=2$) satisfy all three conditions simultaneously — an entire one-parameter family of solutions.

$b=1$ $a=2i$ Re Im

A concrete counterexample: $b=1$ on the unit circle (inner dashed), $a=2i$ on the circle of radius 2 (outer dashed), satisfying all three stated conditions.

Step 5 · Conclusion about the problem as stated

The algebra shows the three given conditions are mutually consistent (they pin down $|a|=2$, $|b|=1$, and a whole family of angle-matched pairs achieving this), and the explicit pair $a=2i,\ b=1$ satisfies every condition exactly. So, as written, with coefficient $2$ in both linear conditions, such complex numbers do exist — the claim "no such $a,b$ exist" is false for this exact statement.

This strongly suggests a transcription issue in the source problem. The classical version of this exercise (which does have a genuine non-existence proof) uses coefficient $1$ instead of $2$:

If $\text{Re}(a)=\text{Im}(b)$ and $\text{Im}(a)=\text{Re}(b)$, the same substitution gives $|a|^2 = y_2^2+x_2^2 = |b|^2$, so $|a|=|b|$. Combined with $|a|-|b|=1$, this forces $0=1$ — a genuine contradiction, proving no such $a,b$ exist in that version.

Result

For the conditions exactly as given (coefficient $2$), the claim is false: solutions exist, e.g.

$$ a = 2i, \qquad b = 1 $$

satisfies $|a|-|b|=1$, $\text{Re}(a)=2\,\text{Im}(b)$, and $\text{Im}(a)=2\,\text{Re}(b)$ simultaneously (verified above), and more generally $a=2\sin\theta+2i\cos\theta$, $b=\cos\theta+i\sin\theta$ works for every $\theta$. If the intended coefficient was $1$ rather than $2$ (i.e. $\text{Re}(a)=\text{Im}(b)$, $\text{Im}(a)=\text{Re}(b)$), then the substitution forces $|a|=|b|$, which together with $|a|-|b|=1$ gives $0=1$ — a genuine contradiction, so no such $a,b$ exist in that version.