Given that z = 5 – 12i express √z in the form a + bi where a, b Є R.
How can I expressed this?
sure,
$\displaystyle
\begin{gathered}
a + ib = \sqrt {a^2 + b^2 } e^{i\arg (a + ib)} \hfill \\
\hfill \\
e^{i\varphi } = \cos \varphi + i\sin \varphi \hfill \\
\hfill \\
\sqrt {xe^{iy} } = \sqrt x e^{i\frac{y}
{2}} \hfill \\
\end{gathered}
$
I also strongly suggest that you read:
Complex number - Wikipedia, the free encyclopedia
Here's another way:
Let $\displaystyle \sqrt{z} = a + ib \Rightarrow z = (a + ib)^2 = (a^2 - b^2) + i(2ab)$
Therefore $\displaystyle 5 - 12i = (a^2 - b^2) + i(2ab)$.
Equate real and imaginary parts:
$\displaystyle 5 = a^2 - b^2$ .... (1)
$\displaystyle -12 = 2ab \Rightarrow -6 = ab \Rightarrow b = -\frac{6}{a}$ .... (2)
Sub (2) into (1):
$\displaystyle 5 = a^2 - \frac{36}{a^2} \Rightarrow a^4 - 5a^2 - 36 = 0 \Rightarrow (a^2 - 9)(a^2 + 4) = 0 \Rightarrow a^2 = 9 \Rightarrow a = \pm 3$.
The solution $\displaystyle a^2 = -4$ is rejected because it gives non-real values of a.
$\displaystyle a = 3 \Rightarrow b = -\frac{6}{3} = -2 \Rightarrow \sqrt{z} = 3 - 2i$.
$\displaystyle a = -3 \Rightarrow b = -\frac{6}{-3} = 2 \Rightarrow \sqrt{z} = -3 + 2i$.
Given how the numbers fall out, I'd say this was the method the question writer had in mind .....