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This is follow up problem to this. I post new question related to previous question since all of the problem were not solved. I have new questions / problems that came after i closed the previous post.

Problem

Compute when $x \in \mathbb{C}$: $$ x^2-4ix-5-i=0 $$ and express output in polar coordinates

Attempt to solve

On last post i solved the equation and got answer:

$$ x=2i \pm \sqrt{i+1} $$

After that i tried to convert this into polar form with little success. Someone provided solution how to express $\sqrt{i+1}$ in a way $\text{Re}$ and $\text{Im}$ parts are separated:

$$ \sqrt{i+1}=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{\sqrt{2}-1}{2}} $$

More about on how this was accomplished is in previous post.

Not in previous post post

Now i tried to solve radius and angle of this complex number. when $x \in \mathbb{C}$

$$ ||x||=\sqrt{\text{Re(x)}^2+\text{Im}(x)^2} $$

$$ \text{Re}(x_1)=\sqrt{\frac{1+\sqrt{2}}{2}} \text{ and } \text{Re}(x_2)=-\sqrt{\frac{1+\sqrt{2}}{2}}$$

$$ \text{Im}(x_1) = 2+ \sqrt{\frac{\sqrt{2}-1}{2}} \text{ and } \text{Im}(x_2) = 2 - \sqrt{\frac{\sqrt{2}-1}{2}}$$

Both complex solutions should have same radius. It is only required to compute radius for one of them.

$$ ||x||=\sqrt{(\sqrt{\frac{1+\sqrt{2}}{2}})^2+( 2+\sqrt{\frac{\sqrt{2}-1}{2}})^2} $$

since $$ (a+b)^2=a^2+2ab+b^2 $$

$$= 2^2+ 2 \cdot 2 \cdot \sqrt{\frac{\sqrt{2}-1}{2}} + (\sqrt{\frac{\sqrt{2}-1}{2}})^2$$

$$= 4+4\sqrt{\frac{\sqrt{2}-1}{2}} + \frac{\sqrt{2}-1}{2} $$

$$||x||= \sqrt{\frac{\sqrt{2}-1}{2}+4+4\sqrt{\frac{\sqrt{2}-1}{2}}+\frac{\sqrt{2}-1}{2}} $$

$$||x||= \sqrt{4\sqrt{\frac{\sqrt{2}-1}{2}}+2\frac{\sqrt{2}-1}{2}+4} $$

$$ \text{let } a = \frac{\sqrt{2}-1}{2}$$

$$ ||x|| = \sqrt{4\sqrt{a}+2a+4} $$

Tried to simplify this expression but doesn't look like it is easily simplified.

Probably have to just stick with this:

$$||x||= \sqrt{4\sqrt{\frac{\sqrt{2}-1}{2}}+2\frac{\sqrt{2}-1}{2}+4} $$

now the argument of complex number

$$ \text{arg}(x)=\arctan(\frac{2+ \sqrt{\frac{\sqrt{2}-1}{2}}}{\sqrt{\frac{1+\sqrt{2}}{2}}}) $$

which again probably doesn't simplify that much.

Now i could express this in polar form which doesn't look that nice:

$$ \sqrt{4\sqrt{\frac{\sqrt{2}-1}{2}}+2\frac{\sqrt{2}-1}{2}+4} \cdot \exp(i\cdot \arctan(\frac{2+ \sqrt{\frac{\sqrt{2}-1}{2}}}{\sqrt{\frac{1+\sqrt{2}}{2}}})) $$

Now problem is this is quite complicated expression and not sure if this is right. If it is possible it would like to have it in much simpler form if this is possible.

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  • $\begingroup$ $$\text{Im}(x_1) = 2i+ i\sqrt{\frac{\sqrt{2}-1}{2}} \text{ and } \text{Im}(x_1) = 2i - i\sqrt{\frac{\sqrt{2}-1}{2}}$$ is without $i$ $\endgroup$ – Ahmad Bazzi Oct 7 '18 at 12:18
  • $\begingroup$ Yes that is mistake @AhmadBazzi $\endgroup$ – Tuki Oct 7 '18 at 12:20
  • $\begingroup$ should be fixed now @AhmadBazzi $\endgroup$ – Tuki Oct 7 '18 at 12:20
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The solution you have provided is correct. The conversion to polar is wrong.

Your mistake is here:

Both complex solutions should have same radius. It is only required to compute radius for one of them. $$||x||=\sqrt{(\sqrt{\frac{1+\sqrt{2}}{2}})^2+( 2+\sqrt{\frac{\sqrt{2}-1}{2}})^2}$$

This is not true. The first radius is the above one, the second is

$$||x||=\sqrt{(\sqrt{\frac{1+\sqrt{2}}{2}})^2+( 2-\sqrt{\frac{\sqrt{2}-1}{2}})^2}$$

Also, there is a calculation mistake, when you arrive here:

$$||x||= \sqrt{4\sqrt{\frac{\sqrt{2}-1}{2}}+2\frac{\sqrt{2}-1}{2}+4}$$

Here is how you do it:

\begin{equation} ||x||=\sqrt{(\sqrt{\frac{1+\sqrt{2}}{2}})^2+( 2+\sqrt{\frac{\sqrt{2}-1}{2}})^2} \end{equation} becomes \begin{equation} ||x||= \sqrt{\frac{1+\sqrt{2}}{2} + 4 + 4\sqrt{\frac{\sqrt{2}-1}{2}} + \frac{\sqrt{2}-1}{2}} \end{equation} that is \begin{equation} ||x||= \sqrt{\sqrt{2} + 4 + 4\sqrt{\frac{\sqrt{2}-1}{2}}} \neq \sqrt{4\sqrt{\frac{\sqrt{2}-1}{2}}+2\frac{\sqrt{2}-1}{2}+4} \end{equation} The first solution has a magnitude given by the above which is not the same magnitude as the second solution. The angle $\theta$ would be the one you have provided, $$\theta=\arctan(\frac{2+ \sqrt{\frac{\sqrt{2}-1}{2}}}{\sqrt{\frac{1+\sqrt{2}}{2}}})$$ So, in polar form, you'd have $$\sqrt{\sqrt{2} + 4 + 4\sqrt{\frac{\sqrt{2}-1}{2}}} \exp(i \arctan(\frac{2+ \sqrt{\frac{\sqrt{2}-1}{2}}}{\sqrt{\frac{1+\sqrt{2}}{2}}}))$$

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  • $\begingroup$ this is correct solution ? $$ \sqrt{4\sqrt{\frac{\sqrt{2}-1}{2}}+2\frac{\sqrt{2}-1}{2}+4} \cdot \exp(i\cdot \arctan(\frac{2+ \sqrt{\frac{\sqrt{2}-1}{2}}}{\sqrt{\frac{1+\sqrt{2}}{2}}})) $$ I am missing the second solution but this one is good ? $\endgroup$ – Tuki Oct 7 '18 at 12:26
  • $\begingroup$ i have edited please see. Also, it is not correct. $\endgroup$ – Ahmad Bazzi Oct 7 '18 at 12:27
  • $\begingroup$ Could you provide the correct solution in polar then ? $\endgroup$ – Tuki Oct 7 '18 at 12:31
  • $\begingroup$ I've pointed out your mistake (i.e now you have the correct magnitude) The rest is just how you did it. $\endgroup$ – Ahmad Bazzi Oct 7 '18 at 12:32
  • $\begingroup$ There is no more "simplified" solution for this problem ? $\endgroup$ – Tuki Oct 7 '18 at 12:33
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$$z^2-4iz-5-i=(z-2i)^2-1-i=0$$

gives

$$z-2i=\pm e^{i\pi/8}=\pm\left(\frac{\sqrt{2+\sqrt2}}2+i\frac{\sqrt{2-\sqrt2}}2\right)$$ by the angle halving formulas.

The polar forms don't seem to be simple.

$$r^2=\|2i\pm e^{i\pi/8}\|^2=4\pm2\sqrt{2-\sqrt2}+1,$$

$$\tan\theta=\frac{4\pm\sqrt{2-\sqrt2}}{\pm\sqrt{2+\sqrt2}}=\frac{\pm4\sqrt{2-\sqrt2}+(2-\sqrt2)}2.$$

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