# Polar form of complex numbers

#### sandy

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

MHF Helper
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 $$\displaystyle \theta = \tan^{-1}\frac{y}{x}$$​

#### sandy

hi

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

#### mr fantastic

MHF Hall of Fame
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

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

MHF Helper
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 $$\displaystyle b = -1$$

Hi ,i tried to do it is this correct?

(-sqrt(3) -i)^7=2^7(cos 7pi/6 + i sin 7pi/6)
$$\displaystyle -\pi < \theta < \pi$$

#### 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

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 $$\displaystyle \theta = \tan^{-1}\frac{y}{x}$$​

#### 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?

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

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

#### 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