Find z s.t. $z+\frac{1}{z}$ is real. Is my solution good?

Question

I must find all z s.t. $z+\frac{1}{z}$ is real. I know that $z = a + bi$ is real when the Imaginary part is 0. So, there we go:

$$z+\frac{1}{z}= \frac{z^2+1}{z}=\frac{(a+bi)^2+1}{a+bi}=\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)}$$ $$\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)} = \frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}$$ $$\frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}=\frac{a^3 +2a^2bi-b^2a+a-ba^2i+2ab^2+b^3i-bi}{(a^2+b^2)}$$ $$2a^2b-ba^2+b^3-b=0$$ $$a^2b+b^3-b=0$$ $$b_1=0$$ $$a^2+b^2-1=0$$ $$b^2=1-a^2$$ $$b_2=\sqrt{1-a^2}$$ $$b_3=-\sqrt{1-a^2}$$

so, the solutions are:

$$z_1=a+0i$$ $$z_1=a+\sqrt{1-a^2}i$$ $$z_1=a-\sqrt{1-a^2}i$$

I am not sure if that should be done like I did it.

$(a+bi)(a-bi) = a^2+b^2$. Check the denominator. – player3236 Nov 01 '20 at 20:17 — player3236, Nov 01 '20 at 20:17
What does s.t. stand for? – A-Level Student Nov 01 '20 at 20:24 — A-Level Student, Nov 01 '20 at 20:24
You can start with $z=r e^{i\phi}$. This should be faster. – user Nov 01 '20 at 20:24 — user, Nov 01 '20 at 20:24

score 3 · Accepted Answer · answered Nov 01 '20 at 20:19

3

I would rather mark (easier to find a mistake) $$z+{1\over z} = c$$ where $c$ is real and now solve $$z^2-zc+1=0$$ It discriminant is $c^2-4$ so $$z_{1,2}= {c\pm \sqrt{c^2-4}\over 2}$$

answered Nov 01 '20 at 20:19

nonuser

90,026

1

Not really, just write $a=c/2$. – nonuser Nov 01 '20 at 20:25
Ok, you are right. Thank you! Deleted my comment because I thought that you edited your answer and now you have different answers. – cocacola Nov 01 '20 at 20:28

Martin R · Answer 2 · 2020-11-01T20:35:37.627

Your solution is correct if you restrict $a \ne 0$ in the case $$ z = a + 0i $$ and $|a| \le 1$ in the case $$ z = a \pm i \sqrt{1-a^2} \, . $$

So for nonzero $z$, $z + 1/z$ is real if $z$ is real or $|z| = 1$.

As alternative solution you can use that a complex number is real if and only if it equal to its complex conjugate: $$ z + \frac 1 z \in \Bbb R \\ \iff z + \frac 1 z = \overline z + \frac{1}{\overline z} \\ \iff z^2 \overline z + \overline z = z \overline z^2 + z \\ \iff (z - \overline z)(|z|^2 - 1) = 0 \, . $$

score 2 · Answer 3 · answered Nov 01 '20 at 20:30

If $z$ is real, you know what happens.

Now if $z$ is not real, the vectors $z$, $1/z$, and $z+1/z$ form a triangle. You would like the complex angle of $z+1/z$ to be $0$.

The vectors $z$ and $1/z$ have complex angles that are negative of each other. So now this triangle has two equal angles with $z+1/z$ representing the edge between them. It's an isosceles triangle, and it follows that the magnitude of $z$ equals that of $1/z$.

So the only $z$ for which this can hold are the $z$ on the unit circle. And you can then verify that for all such complex numbers, the sum is real: $e^{it}+e^{-it}=2\cos(t)$.

score 1 · Answer 4 · answered Nov 01 '20 at 21:20

Your method works fine, however it's rather algebra-heavy which can lead to accidental mistakes. I would note that $$z+\frac 1z \in \Bbb R \iff \mathfrak{Im}(z+\frac 1z)=\mathfrak{Im}(z)+\mathfrak{Im}(\frac 1z)=0$$

Then, it is easily seen that $$\mathfrak{Im}(\frac 1z)=\frac{-\mathfrak{Im}(z)}{|z|^2}$$ and so we are solving the equation: $$\mathfrak{Im}(z)\bigg(1-\frac{1}{|z|^2}\bigg)=0$$ which has solutions when $\mathfrak{Im}(z)=0$ (i.e. $z\in\Bbb R)$ or $$|z|^2=1\implies |z|=1\implies \mathfrak{Im}^2(z)=1-\mathfrak{Re}^2(z)\to z=t\pm i\sqrt{1-t^2}$$ which is what you got.

Find z s.t. $z+\frac{1}{z}$ is real. Is my solution good?

4 Answers4