Prove there is no complex z such that $|z|=|z + i\sqrt5| = 1$

Question

This is a question in introduction to pure mathematics. I am pretty sure I am close to the answer but I can't quite decide why this proves that there is no complex numbers:

$$|z| = |z + i√5| = 1$$ $$\sqrt{cos²Θ + (isinΘ + i\sqrt5)²} = \sqrt1$$ $$cos²Θ - sin²Θ - 2\sqrt5sinΘ - 5 = 1$$ $$cos²Θ - (1 - cos²Θ) - 2\sqrt5sinΘ - 5 = 1$$ $$2cos²Θ - 2√5sinΘ - 5 = 0$$

I can see are that there is no imaginary component left in the equation and there is no way to equate it to $cos\theta + isin\theta$.

Answer 1

Consider a simple geometric interpretation: if $z$ is a point on the unit circle in the complex plane, and we require $z+i\sqrt{5}$ to also be on this circle, this is obviously impossible since the diameter of the circle is $2$: for any line drawn through the circle, the points of intersection of that line with the circle cannot possibly be more than a distance of $2$ apart, yet $|i \sqrt{5}| = \sqrt{5} > 2$.

Answer 2

$$2\cos^2\theta-2\sqrt5\sin\theta-5=0\iff 2\sin^2\theta+2\sqrt5\sin\theta+3=0$$

The discriminant of the above quadratic in $\;\sin\theta\;$ is

$$\Delta=20-24<0\implies\; \text{there is no real solution}\;\ldots$$

Answer 3

Hint: Delete all $i$'s in your second equation. At end of day, you'll just need $|\sin \theta |\leq 1$ for all $\theta$.

Answer 4

Another way:

If $\displaystyle|z|=|z-id|$ and $z=a+ib$ where $d\ne0, a,b$ are real

$$a^2+b^2=a^2+(b-d)^2\iff 2bd=d^2\iff b=\frac d2$$

So, $\displaystyle a^2+b^2=a^2+\left(\frac d2\right)^2\ge\frac{d^2}4$

Here $\displaystyle d=-\sqrt5\implies d^2=5,a^2+b^2\ge\frac{5}4>1$