2
$\begingroup$

Problem

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

Attempt to solve

Solving this equation with quadratic formula:

$$ x=\frac{4i \pm \sqrt{(-4i)^2-4\cdot (-5-i)}}{2} $$ $$x= \frac{4i \pm \sqrt{4(i+1)}}{2} $$ $$ x = \frac{4i \pm 2\sqrt{i+1}}{2} $$ $$ x = 2i \pm \sqrt{i+1} $$

I can transform cartesian complex numbers to polar with eulers formula: when $z \in \mathbb{C}$

$$ z=re^{i\theta} $$

then: $$ r=|z|=\sqrt{(\text{Re(z)})^2+(\text{Im(z)})^2} $$ $$ \text{arg}(x)=\theta = \arctan{\frac{\text{Im}(z)}{\text{Re}(z)}} $$

Plugging in values after this computation would give us our complex in number in $(r,\theta)$ polar coordinates from $(\text{Re},\text{Im})$ cartesian coordinates.

Only problem is how do i convert complex number of form $$ z=2i+\sqrt{i+1} $$ to polar since i don't know how to separate this into imaginary and real parts. How do you compute $\text{Re}(z)$ and $\text{Im}(z)$

$\endgroup$
  • $\begingroup$ SInce $1+i=\sqrt{2}\exp\frac{\pi i}{4}$, $\sqrt{1+i}=\sqrt[4]{2}(\cos\frac{\pi i}{8}+i\sin\frac{\pi i}{8})$. You can thereby get the real and imaginary parts of $z$ and convert to polar coordinates. $\endgroup$ – J.G. Oct 6 '18 at 21:10
  • 1
    $\begingroup$ where does $1+i=\sqrt{2}\exp(\frac{\pi i}{4})$ come from ? @J.G. $\endgroup$ – Tuki Oct 6 '18 at 21:12
  • $\begingroup$ derived from $e^{i\pi}+1=0$ i suppose ? $\endgroup$ – Tuki Oct 6 '18 at 21:13
2
$\begingroup$

Let $a,b\in\mathbb{R}$ so that $$\sqrt{i+1} = a+bi$$ $$ i+1 = a^2 -b^2 +2abi $$

Equating real and imaginary parts, we have

$$2ab = 1$$

$$a^2 -b^2 = 1$$


Now we solve for $(a,b)$. $$ \begin{align*} b &= \frac{1}{2a}\\\\ \implies \,\,\, a^2 - \left(\frac{1}{2a}\right)^2 &= 1 \\\\ a^2 &= 1 + \frac{1}{4a^2}\\\\ 4a^4 &= 4a^2 + 1\\\\ 4a^4 - 4a^2 -1 &= 0 \\\\ \end{align*} $$

This is a quadratic in $a^2$ (it's also a quadratic in $2a^2$, if you prefer!), so we use the quadratic formula:

$$a^2 = \frac{4 \pm \sqrt{16-4(4)(-1)}}{2(4)}$$

$$a^2 = \frac{1 \pm \sqrt{2}}{2}$$

Here we note that $a$ is real, so $a^2>0$, and we discard the negative case:

$$a^2 = \frac{1 + \sqrt{2}}{2}$$

$$a = \pm \sqrt{\frac{1 + \sqrt{2}}{2}}$$

$$ b = \frac{1}{2a} = \pm \sqrt{\frac{\sqrt{2}-1}{2}}$$


This gives what you can call the principal root:

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

As well as the negation of it:

$$-\sqrt{i+1} = -\sqrt{\frac{1 + \sqrt{2}}{2}} + i\left(-\sqrt{\frac{\sqrt{2}-1}{2}}\right) $$


Finally, substituting either of these into your expression $$z=2i \pm \sqrt{i+1}$$ will give you $\text{Re}(z)$ and $\text{Im}(z)$.

At that point, as you noted in your question, conversion to polar coordinates is straightforward.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.