# Polar form of complex numbers

• May 13th 2010, 01:52 PM
sandy
Polar form of complex numbers
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
• May 13th 2010, 02:08 PM
pickslides
Start by using this, where $\displaystyle z = x+yi$ then $\displaystyle (x+yi)^n = z^n = r^n(\cos n\theta +i\sin n\theta)$

given $\displaystyle r = \sqrt{x^2+y^2}$ and $\displaystyle \theta = \tan^{-1}\frac{y}{x}$
• May 13th 2010, 07:51 PM
sandy
hi
i did start with that but im notgetting the right answer i dont know what im doing wrong
• May 13th 2010, 08:00 PM
mr fantastic
Quote:

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.
• May 14th 2010, 03:42 PM
sandy
Quote:

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
• May 14th 2010, 03:56 PM
pickslides
Quote:

Originally Posted by sandy

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

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

Quote:

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)

$\displaystyle -\pi < \theta < \pi$
• May 14th 2010, 03:56 PM
sandy
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

Quote:

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

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

• May 14th 2010, 04:00 PM
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?

Quote:

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

$\displaystyle -\pi < \theta < \pi$

• May 14th 2010, 04:12 PM
sandy
Quote:

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