Problem in Complex Number

**geton** · January 28th 2008, 03:27 AM

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

How can I expressed this?

**Peritus** · January 28th 2008, 03:36 AM

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

**geton** · January 28th 2008, 04:01 AM

Can you explain more?

**Peritus** · January 28th 2008, 04:07 AM

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

**mr fantastic** · January 28th 2008, 04:13 AM

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 .....

**geton** · January 28th 2008, 04:30 AM

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

**geton** · January 28th 2008, 04:33 AM

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

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