Polar form of complex numbers

**sandy** · May 13th 2010, 01:52 PM

hi guys , i really need help in these two questions :

Express each of the following in the form

a + ib and also in the polar form r\µ,
where the angle is the principal value.
(a) (-sqrt(3)- i )^7

(b)(1 + i)^3(sqrt(3) + i)^3

**pickslides** · May 13th 2010, 02:08 PM

Start by using this, where $z = x+yi$ then $(x+yi)^n = z^n = r^n(\cos n\theta +i\sin n\theta)$

given $r = \sqrt{x^2+y^2}$ and $\theta = \tan^{-1}\frac{y}{x}$

**sandy** · May 13th 2010, 07:51 PM

i did start with that but im notgetting the right answer i dont know what im doing wrong

**mr fantastic** · May 13th 2010, 08:00 PM

Originally Posted by sandy

i did start with that but im notgetting the right answer i dont know what im doing wrong

We won't know either unless you show all your working.

**sandy** · May 14th 2010, 03:42 PM

Originally Posted by mr fantastic

We won't know either unless you show all your working.

Hi ,i tried to do it is this correct?

(-sqrt(3) -i)^7=2^7(cos 7pi/6 + i sin 7pi/6)

and how do we do it with the form a+ib

**pickslides** · May 14th 2010, 03:56 PM

Originally Posted by sandy

and how do we do it with the form a+ib

That's exactly what we are doing. you have $-\sqrt{3}-i$ giving $a = -\sqrt{3}$ and $b = -1$

Originally Posted by sandy

Hi ,i tried to do it is this correct?

(-sqrt(3) -i)^7=2^7(cos 7pi/6 + i sin 7pi/6)

$-\pi < \theta < \pi$

**sandy** · May 14th 2010, 03:56 PM

Hi ,i tried to do it is this correct?

(-sqrt(3) -i)^7=2^7(cos 7pi/6 + i sin 7pi/6)

and how do we do it with the form a+ib

Originally Posted by pickslides

Start by using this, where $z = x+yi$ then $(x+yi)^n = z^n = r^n(\cos n\theta +i\sin n\theta)$

given $r = \sqrt{x^2+y^2}$ and $\theta = \tan^{-1}\frac{y}{x}$

**sandy** · May 14th 2010, 04:00 PM

So do we expand a+ib to the power of 7 by to the power of 2 multiplied by 3 and then multiply that by a+ib again?

Originally Posted by pickslides

That's exactly what we are doing. you have $-\sqrt{3}-i$ giving $a = -\sqrt{3}$ and $b = -1$

$-\pi < \theta < \pi$

**sandy** · May 14th 2010, 04:12 PM

Originally Posted by sandy

So do we expand a+ib to the power of 7 by to the power of 2 multiplied by 3 and then multiply that by a+ib again?

i got it !!
z^7= 64*sqrt(3) - i64

Thread: Polar form of complex numbers