Polar form of complex numbers

Apr 2010
43
0
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
Sep 2008
5,237
1,625
Melbourne
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 \(\displaystyle \theta = \tan^{-1}\frac{y}{x}\)​
 
Apr 2010
43
0
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
Dec 2007
16,948
6,768
Zeitgeist
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.
 
Apr 2010
43
0
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
Sep 2008
5,237
1,625
Melbourne
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)
\(\displaystyle -\pi < \theta < \pi\)
 
Apr 2010
43
0
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 \(\displaystyle \theta = \tan^{-1}\frac{y}{x}\)​
 
Apr 2010
43
0
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 \(\displaystyle b = -1\)




\(\displaystyle -\pi < \theta < \pi\)
 
Apr 2010
43
0
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