S sandy Apr 2010 43 0 May 13, 2010 #1 hi guys , i really need help in these two questions : Express each of the following in the forma + 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

hi guys , i really need help in these two questions : Express each of the following in the forma + 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 May 13, 2010 #2 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 \(\displaystyle \theta = \tan^{-1}\frac{y}{x}\)

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 \(\displaystyle \theta = \tan^{-1}\frac{y}{x}\)

S sandy Apr 2010 43 0 May 13, 2010 #3 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 May 13, 2010 #4 sandy said: i did start with that but im notgetting the right answer i dont know what im doing wrong Click to expand... We won't know either unless you show all your working.

sandy said: i did start with that but im notgetting the right answer i dont know what im doing wrong Click to expand... We won't know either unless you show all your working.

S sandy Apr 2010 43 0 May 14, 2010 #5 mr fantastic said: We won't know either unless you show all your working. Click to expand... 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

mr fantastic said: We won't know either unless you show all your working. Click to expand... 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 May 14, 2010 #6 sandy said: and how do we do it with the form a+ib Click to expand... That's exactly what we are doing. you have \(\displaystyle -\sqrt{3}-i\) giving \(\displaystyle a = -\sqrt{3}\) and \(\displaystyle b = -1\) sandy said: Hi ,i tried to do it is this correct? (-sqrt(3) -i)^7=2^7(cos 7pi/6 + i sin 7pi/6) Click to expand... \(\displaystyle -\pi < \theta < \pi\)

sandy said: and how do we do it with the form a+ib Click to expand... That's exactly what we are doing. you have \(\displaystyle -\sqrt{3}-i\) giving \(\displaystyle a = -\sqrt{3}\) and \(\displaystyle b = -1\) sandy said: Hi ,i tried to do it is this correct? (-sqrt(3) -i)^7=2^7(cos 7pi/6 + i sin 7pi/6) Click to expand... \(\displaystyle -\pi < \theta < \pi\)

S sandy Apr 2010 43 0 May 14, 2010 #7 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 said: 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 \(\displaystyle \theta = \tan^{-1}\frac{y}{x}\) Click to expand...

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 said: 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 \(\displaystyle \theta = \tan^{-1}\frac{y}{x}\) Click to expand...

S sandy Apr 2010 43 0 May 14, 2010 #8 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? pickslides said: That's exactly what we are doing. you have \(\displaystyle -\sqrt{3}-i\) giving \(\displaystyle a = -\sqrt{3}\) and \(\displaystyle b = -1\) \(\displaystyle -\pi < \theta < \pi\) Click to expand...

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? pickslides said: That's exactly what we are doing. you have \(\displaystyle -\sqrt{3}-i\) giving \(\displaystyle a = -\sqrt{3}\) and \(\displaystyle b = -1\) \(\displaystyle -\pi < \theta < \pi\) Click to expand...

S sandy Apr 2010 43 0 May 14, 2010 #9 sandy said: 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? Click to expand... i got it !! z^7= 64*sqrt(3) - i64

sandy said: 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? Click to expand... i got it !! z^7= 64*sqrt(3) - i64