solutions in polar form?

[cazza90]

So what are you supposed to do for questions when factors of z equal zero, and you are asked to solve and put your solutions in polar form?
Eg find solutions of ((z^3) + i)((z^2) + z + 1) = 0 amd give roots in polar form?

 

[Deleted member 4993]

So what are you supposed to do for questions when factors of z equal zero, and you are asked to solve and put your solutions in polar form?
Eg find solutions of ((z^3) + i)((z^2) + z + 1) = 0 amd give roots in polar form?

What is exactly meant by that!

One of the sets of solutions is:

z[sup:2z708y55]3[/sup:2z708y55] + i = 0

z[sup:2z708y55]3[/sup:2z708y55] = - i

z[sup:2z708y55]3[/sup:2z708y55] = e [sup:2z708y55]i(3/2 + 2n)?[/sup:2z708y55]

Now continue.....

 

[tkhunny]

You simply do it.

You have z^3 = -i.

What are the three cube roots of -i? DeMoivre's formula can be quite useful, of course.

\(\displaystyle i = cis\left(\frac{\pi}{2}\right)\)

Then \(\displaystyle \frac{1}{2}(-\sqrt{3}-i) = cis\left(\frac{7\pi}{6}\right)\)

Keep going around the circle. You'll find the third one.

 

[Deleted member 4993]

So what are you supposed to do for questions when factors of z equal zero, and you are asked to solve and put your solutions in polar form?
Eg find solutions of ((z^3) + i)((z^2) + z + 1) = 0 amd give roots in polar form?

You'll get two sets of roots from z[sup:1k241dk5]2[/sup:1k241dk5] + z + 1 = 0

 

[iambryanatkinson]

for z[sup:2vfm2h1o]2[/sup:2vfm2h1o]+ z + 1 = 0 , we get

\(\displaystyle z = {{-1 \pm \sqrt{1 - 4}} \over 2}\)
\(\displaystyle z = -{1 \over 2} + i{\sqrt{3} \over 2} , -{1 \over 2} - i{\sqrt{3} \over 2}\)

in polar form,
\(\displaystyle z = e^{5\pi \over 6}, e^{7\pi \over 6}\)


add this to Subhotosh Khan's answer and we get completed answer