Determine the sq root of complex number

**zorro** · December 28th 2009, 11:48 AM

Question : Determine the sq root of $(3+4i) , \ \ i=\sqrt{-1}$

**bigwave** · December 28th 2009, 12:00 PM

Originally Posted by zorro

Question : Determine the sq root of $(3+4i) , \ \ i=\sqrt{-1}$

this is the same as

$ 4 + 4i - 1 $
which factors to

$ (2+i)(2+i)$

or $(2+i)^2$

which $\sqrt{(2+i)^2} = \pm(2+i )$

**Plato** · December 28th 2009, 12:01 PM

Originally Posted by zorro

Question : Determine the sq root of $(3+4i) , \ \ i=\sqrt{-1}$

zorro, Do you ever try to do any of the many problems that you post?
I have not seen any evidence of your work.

**zorro** · December 28th 2009, 12:25 PM

But how should i find the amplitute $\theta$ = $tan^{-1} \left( \frac{y}{x} \right)$ = $tan^{-1} \left( \frac{1}{2} \right)$ = ?

**bigwave** · December 28th 2009, 12:42 PM

ok latex error went away

Amplitude is the magnitude of change in an oscillating variable

**zorro** · December 28th 2009, 12:43 PM

No i want to know what would th amplitute be .........there is no problem with the latex

**Abu-Khalil** · December 28th 2009, 01:52 PM

$\sqrt{\left(3+4i\right)}=\sqrt{5e^{i\theta+2\pi}}= \sqrt{5}e^{i\frac{\theta}{2}+\pi},\theta=\arctan\l eft(\frac{3}{4}\right).$

**zorro** · December 28th 2009, 02:07 PM

i know that mite i want ot know the value of $\theta$ , i need to find the cos and sin value with it ......

**Abu-Khalil** · December 28th 2009, 02:16 PM

You have a triangle rectangle which sides are $3$ and $4$ . Hipotenuse?

**zorro** · December 28th 2009, 02:32 PM

Originally Posted by Abu-Khalil

You have a triangle rectangle which sides are $3$ and $4$ . Hipotenuse?

i am not getting it mite ...........the reason i am not getting is because i need the value of $\theta$ ..............and ur giving me clues which i can get it ..........

**Abu-Khalil** · December 28th 2009, 02:47 PM

Originally Posted by zorro

i am not getting it mite ...........the reason i am not getting is because i need the value of $\theta$ ..............and ur giving me clues which i can get it ..........

You don't need the value of $\theta=\arctan\left(\frac{3}{4}\right)$ , 'cause that means $\cos\theta=\frac{4}{5},\sin\theta=\frac{3}{5}$ .

**zorro** · December 28th 2009, 03:25 PM

No u are not getting it i want to express this in polar form ........

**Also sprach Zarathustra** · December 28th 2009, 03:46 PM

3+4i=5(cos(38.86...+2pi*k)* + isin(36.86...+2pi*k)*)=5cis(36.86... +2pi*k)*

square root of above can be calculate with de Moivre formula.
De Moivre's formula - Wikipedia, the free encyclopedia

**Prove It** · December 28th 2009, 03:55 PM

Originally Posted by bigwave

this is the same as

$ 4 + 4i - 1 $
which factors to

$ (2+i)(2+i)$

or $(2+i)^2$

which $\sqrt{(2+i)^2} = 2+i$

Actually it's $2 + i$ or $-(2 + i)$ .

**zorro** · December 28th 2009, 05:14 PM

could anybody tell me what is the polar form of this equation ............

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