Multiplication of real and complex radicals

Question

If I have, for example, the product

$\sqrt{7+\sqrt{22}}\sqrt[3]{38+i\sqrt{6}} $

Can I perform the multiplication or this cannot be done and only remains to leave the product in this form?

Answer 1

Your question is ambiguous, because there is no such thing as a radical for complex numbers. It's not well defined, because you can't choose a specific complex number for your function (on $\mathbb R$ we choose the positive one).

So if you define $\sqrt.$ properly, may be you could perform your product, but it seems a lot of trouble.

Answer 2

$\sqrt{7+\sqrt{22}}\sqrt[3]{38+\sqrt6i}= \sqrt[6]{(7+\sqrt{22})^3}\sqrt[6]{(38+\sqrt6i)^2}= \sqrt[6]{(7+\sqrt{22})^3(38+\sqrt6i)^2}= \sqrt[6]{1157590+243022\sqrt{22}+(61180\sqrt6+25688\sqrt{33})i}$

EDIT:
Other possible valid results are the above result multiplied by $\left(\cos{\frac{k\pi}3}+i\sin{\frac{k\pi}3}\right)\text{, where}\;k=\{1\dots 5\}$

Answer 3

As noted in another answer, the radical isn't an easy thing to understand for complex numbers. However, we can simplify it by taking the "principle branch".

For a complex number, we get that $a+bi=z=|z|e^{i(\theta+2k\pi)}$ for any integer $k$ (see below), where $|z|=\sqrt{a^2+b^2}$ is the distance from the origin in the complex plane and $\theta=\arctan(b/a)$ is the angle with the positive real axis. Here we will let $k=0$ only for simplicity. Hence $z^{1/n}=|z|^{1/n}e^{i\theta/n}.$

Now we will use Euler's identity $e^{i\theta}=\cos\theta+i\sin\theta.$ (This is a well know fact from higher mathematics.)

$$z^{1/n}=|z|^{1/n}\left(\cos(\theta/n)+i\sin(\theta/n)\right).$$

So now, you just have to turn your complex numbers into their respective polar forms, take the "principle roots" and then multiply/distribute as per usual algebraic operations.

You example has $z=38+i\sqrt{6}.$ This gives $|z|=\sqrt{38^2+6}=\sqrt{1450}=5\sqrt{2\cdot29}$ and $\theta=\arctan(\sqrt{6}/38).$

Thus $$z^{1/3}=5^{1/3}58^{1/6}\left(\cos\left(\frac{\arctan(\sqrt{6}/38)}{3}\right)+i\sin\left(\frac{\arctan(\sqrt{6}/38)}{3}\right)\right).$$

Now just multiply this by $\sqrt{7+\sqrt{22}}$ to get

$$\left(7+\sqrt{22}\right)^{1/2}\cdot 5^{1/3}\cdot 58^{1/6}\left(\cos\left(\frac{\arctan(\sqrt{6}/38)}{3}\right)+i\sin\left(\frac{\arctan(\sqrt{6}/38)}{3}\right)\right).$$

This can now be evaluated in a calculator in order to get a decimal approximation: $11.5003+0.2468i.$

Of course, the general solution is

$$\left(7+\sqrt{22}\right)^{1/2}\cdot 5^{1/3}\cdot 58^{1/6}\left(\cos\left(\frac{\arctan(\sqrt{6}/38)+2k\pi}{3}\right)+i\sin\left(\frac{\arctan(\sqrt{6}/38)+2k\pi}{3}\right)\right)$$

for $k=0,1,2.$ Since we took a cube root, we have to find three distinct roots.

Why the $2k\pi$?

When you think of a complex number as a point in the complex plane, you should realize it can be characterized by its distance from the origin and the angle it makes with the positive real axis. So if we rotate around the origin by $360^\circ$, then we end up back at the same point. This is where the $2k\pi=360^\circ \cdot k$ comes from.