useabe solution to the calculation sqrt(3-0.5i)

[sward071]

Hey being wrecking my head trying to come up with a useabe solution to the calculation

sqrt(3-0.5i)

so that i can use it for analysis but cant get a answer.

please can someone tell me the steps involved or what I should expect as a answer

 

[stapel]

trying to come up with a useabe solution to the calculation

sqrt(3-0.5i)

What do you mean by "a usable solution"? By "calculation", do you mean "decimal approximation"? If so, are you saying that "i" is a variable of some sort, rather than the imaginary? If so, what is the value of "i"?

It might help if you provided the context in which this expression arose. Thank you!

 

[Ishuda]

Hey being wrecking my head trying to come up with a useabe solution to the calculation

sqrt(3-0.5i)

so that i can use it for analysis but cant get a answer.

please can someone tell me the steps involved or what I should expect as a answer

Assuming [there's that word again] what you mean is the square root of the complex number 3 - 0.5 i, I generally turn it into an r-\(\displaystyle \theta]\) number

c = a + b i = r \(\displaystyle e^{i\theta}= r cos(\theta) + r i sin(\theta)\)
where
r = \(\displaystyle \sqrt{a^2+b^2}\)
and
\(\displaystyle \theta = atan2(b, a)\)

Once you have the answer convert back to Cartesian co-ordinates again if wanted. See
http://en.wikipedia.org/wiki/Atan2
for an example of the atan2 function.

 

[soroban]

Hello, sward071!

I assume you want: \(\displaystyle \:\sqrt{3 - \frac{1}{2}i}\)

Let \(\displaystyle a\, +\, bi \:=\:\sqrt{3\,-\,\frac{1}{2}i\,}\)

where \(\displaystyle a\) and \(\displaystyle b\) are real, and \(\displaystyle a\,>\,0.\) Then:

\(\displaystyle (a\,+\,bi)^2 \;=\;3\, -\, \frac{1}{2}i\)

\(\displaystyle (a^2\,-\,b^2)\, +\, (2ab)i \;=\;3 \,-\, \frac{1}{2}i\)

We have:

\(\displaystyle \begin{Bmatrix}a^2-b^2 &=&3 & [1]
\\ 2ab &=& -\frac{1}{2} & [2] \end{Bmatrix}\)

From [2]: \(\displaystyle b \:=\:-\dfrac{1}{4a}\;\;[3]\)

Substitute into [1]:

\(\displaystyle a^2\,- \,\dfrac{1}{16a^2}\;=\;3\)

And we have:

\(\displaystyle 16a^4\, - \,48a^2 \,\pm\, 1 \;=\;0\)

Quadratic Formula:

\(\displaystyle a^2\;=\;\dfrac{48\, \pm\,\sqrt{2368\,}}{32} \;=\; \dfrac{48 \,\pm \,8\sqrt{37\,}}{32}\)

\(\displaystyle a^2 \;=\;\dfrac{6\, \pm\,\sqrt{37\,}}{4} \quad\,\Rightarrow\,\quad a \;=\;\dfrac{\sqrt{\sqrt{37\,}\,+\,6}}{2} \)

Substitute into [3]:

\(\displaystyle b \;=\;-\dfrac{\sqrt{\sqrt{37\,}\, -\, 6}}{2} \)

Therefore: \(\displaystyle a + bi \;=....\)