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 $\text{Im}(z)$

  • $\begingroup$ 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, 2018 at 21:10
  • 1
    $\begingroup$ where does $1+i=\sqrt{2}\exp(\frac{\pi i}{4})$ come from ? @J.G. $\endgroup$
    – Tuki
    Oct 6, 2018 at 21:12
  • $\begingroup$ derived from $e^{i\pi}+1=0$ i suppose ? $\endgroup$
    – Tuki
    Oct 6, 2018 at 21:13

1 Answer 1


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.


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