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Mathematics 19 Online
OpenStudy (anonymous):

. Given an equation zsq+ 1/zsq = 1 , solve for z in polar form

OpenStudy (anonymous):

\[z ^{2} + 1/ z ^{2} = 1\]

OpenStudy (dumbcow):

oh that was z^2 lol

OpenStudy (anonymous):

Do you know if Z is a complex number or is it just a random variable like X?

OpenStudy (anonymous):

Its a random variable like X

OpenStudy (dumbcow):

i get \[z^{2} = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i\]

OpenStudy (dumbcow):

which in polar form is \[z^{2} = \cos \pi/3 +i \sin \pi/3\]

OpenStudy (anonymous):

The answer is \[1 Angle \pi/6\]

OpenStudy (dumbcow):

hmm z = pi/6 is the answer

OpenStudy (dumbcow):

oh wait, its that one rule i forget the name \[z = [\cos \pi/3 + i \sin \pi/3]^{1/2}\] \[z = (1/2)(\cos \pi/6 + i \sin \pi/6)\]

OpenStudy (dumbcow):

does that look right??

OpenStudy (anonymous):

Looks good to me lol.

OpenStudy (dumbcow):

its De Moivres theorem i did a little wrong though the 1/2 should not be out front \[z =(\cos \pi/6 + i \sin \pi/6)\]

OpenStudy (anonymous):

ouhkk, thanks

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