1

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 $cos\theta + isin\theta$.

Matt Martineau
  • 151
  • Yes sorry, should be fixed now, is there any way to preview title? – Matt Martineau Mar 21 '14 at 16:06
  • I can't tell...but it is fine, now. – DonAntonio Mar 21 '14 at 16:08
  • 1
    The magnitude of a complex number is real (and nonnegative), so it wouldn't have any imaginary component to begin with. By writing $z=\cos\theta+i\sin\theta,$ you've already imposed $\lvert z\rvert=1,$ so you don't need to do this again. Also, the magnitude of $x+iy$ is $\sqrt{x^2+y^2},$ not $\sqrt{x^2+(iy)^2}.$ – Will Orrick Mar 21 '14 at 16:08
  • 7
    Because $|z|=1$ and $|z+i\sqrt{5}|=1$ are circles of radius $1$, and their centers are $\sqrt{5}>2$ units apart, they have no point in common. – mjqxxxx Mar 21 '14 at 16:09

4 Answers4

4

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 $|i \sqrt{5}| = \sqrt{5} > 2$.

heropup
  • 135,869
  • This is why I clicked on the question on the home page. If this answer wasn't already there I would have posted. Good job. +1 Cheers. – Guy Mar 21 '14 at 16:21
  • This is great, thank you. I have not been considering complex numbers as circles on the complex plane but this is very intuitive. – Matt Martineau Mar 22 '14 at 06:35
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

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

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
2

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

ir7
  • 6,249
1

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

lab bhattacharjee
  • 274,582