# Convert complex number to polar coordinates

## 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 $$\text{Im}(z)$$

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

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.