# Thread: Problem in Complex Number

1. ## Problem in Complex Number

Given that z = 5 – 12i express √z in the form a + bi where a, b Є R.

How can I expressed this?

2. The easyest way of doing this is convert the complex number to polar form then take the square root and convert back to cartesian.

3. Can you explain more?

4. Originally Posted by geton
Can you explain more?
sure,
$
\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

5. Originally Posted by geton
Given that z = 5 – 12i express √z in the form a + bi where a, b Є R.

How can I expressed this?
Here's another way:

Let $\sqrt{z} = a + ib \Rightarrow z = (a + ib)^2 = (a^2 - b^2) + i(2ab)$

Therefore $5 - 12i = (a^2 - b^2) + i(2ab)$.

Equate real and imaginary parts:

$5 = a^2 - b^2$ .... (1)

$-12 = 2ab \Rightarrow -6 = ab \Rightarrow b = -\frac{6}{a}$ .... (2)

Sub (2) into (1):

$5 = a^2 - \frac{36}{a^2} \Rightarrow a^4 - 5a^2 - 36 = 0 \Rightarrow (a^2 - 9)(a^2 + 4) = 0 \Rightarrow a^2 = 9 \Rightarrow a = \pm 3$.

The solution $a^2 = -4$ is rejected because it gives non-real values of a.

$a = 3 \Rightarrow b = -\frac{6}{3} = -2 \Rightarrow \sqrt{z} = 3 - 2i$.

$a = -3 \Rightarrow b = -\frac{6}{-3} = 2 \Rightarrow \sqrt{z} = -3 + 2i$.

Given how the numbers fall out, I'd say this was the method the question writer had in mind .....

6. Originally Posted by Peritus
sure,
$
\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
Thank you so much. I got it

7. Originally Posted by mr fantastic
Here's another way:

Let $\sqrt{z} = a + ib \Rightarrow z = (a + ib)^2 = (a^2 - b^2) + i(2ab)$

Therefore $5 - 12i = (a^2 - b^2) + i(2ab)$.

Equate real and imaginary parts:

$5 = a^2 - b^2$ .... (1)

$-12 = 2ab \Rightarrow -6 = ab \Rightarrow b = -\frac{6}{a}$ .... (2)

Sub (2) into (1):

$5 = a^2 - \frac{36}{a^2} \Rightarrow a^4 - 5a^2 - 36 = 0 \Rightarrow (a^2 - 9)(a^2 + 4) = 0 \Rightarrow a^2 = 9 \Rightarrow a = \pm 3$.

The solution $a^2 = -4$ is rejected because it gives non-real values of a.

$a = 3 \Rightarrow b = -\frac{6}{3} = -2 \Rightarrow \sqrt{z} = 3 - 2i$.

$a = -3 \Rightarrow b = -\frac{6}{-3} = 2 \Rightarrow \sqrt{z} = -3 + 2i$.

Given how the numbers fall out, I'd say this was the method the question writer had in mind .....
Yes I forgot that. Thank you