Square root from (6i-8)

Now let's do simple practice with the calculation of square root from imaginary numbers. Let's calculate \(\sqrt{6i - 8} = ?\).

It is important to spot, that this is a bit unusual order in the task. Actually, in the classical format it should be: \(\sqrt{-8 + 6i} = ?\), and you need to be careful, using the standard equation. \(a = -8; b = 6\)

So, let's apply classical formula:

\[\sqrt{a + ib} = \pm(\sqrt{\frac{\left|z\right| + a}{2}} + i\frac{b}{\left|b\right|}\sqrt{\frac{\left|z\right| - a}{2}})\]

First of all, let's estimate \(\left|z\right|\) and \(sign(b)\):

\[\left|z\right| = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\]

\[sign(b) = \frac{b}{\left|b\right|} = \frac{6}{\left|6\right|} = 1\]

and now le's calculate the absolute values of a real and imaginary parts:

\[\sqrt{\frac{\left|z\right| + a}{2}} = \sqrt{\frac{10 + (-8)}{2}} = 1 \\\sqrt{\frac{\left|z\right| - a}{2}} = \sqrt{\frac{10 - (-8)}{2}} = \sqrt{9} = 3\]

Therefore, the final answer will be:

\[\sqrt{6i - 8} = \pm(1 + 3i)\]

For more details and some comments please check this video:

Published: 2023-05-08 22:19:54