Find all complex numbers z such that Re(z^3)<0 PLEEASSE HELP (1 Viewer)

[adnan91]

Full question: Find all complex numbers z such that Re(z^3)<0 and show the

solutions graphically in the complex plane.

It says hint: this problem is best handled in polar coordinations.

Ive made little progress in it and i have no idea what to do. HELP ME!!

thanks in advnce

 

[adnan91]

umpbay

 

[lolokay]

if z = Rcis@
then z³ = R³cis[3@]

for the condition to hold, what ranges of values can 3@ have? so what range does @ have?

 

[Trebla]

Why is this in the General Maths forum? lol

Basically you want R³cos @ < 0, since R is always positive you hence require cos 3@ < 0. Once you found general solutions of @, you can then determine the region.

 

[xxstef]

itd be interesting if any general maths student could help you on this, lol.

 

[PC]

It says hint: this problem is best handled in polar coordinations.

Does that mean you have to be in Antarctica to answer the question?

 

[tommykins]

Does that mean you have to be in Antarctica to answer the question?

oh my lol.

 

[EvoRevolution]

This is how far i have gotten in to Antarctica..


Re(z^3)<0
Re(r^3cis(3@))<0 if z = rcis@
Re(r^3cos3@ + isin@)<0
r^3cos3@<0


then somethin else

 

[cutemouse]

General Maths? Hahahah

Just let z=x+iy and use the expansion (a+b)³=a³+3ab²+3a²b+b³

And then gather all the real parts.

 

[youngminii]

Polar form is rcis@, which is much neater than cartesian form for this question..
Man so much help has been given in this thread ==;;
Re(z^3) < 0
Let z = rcis@
Re(r^3cis3@) < 0
r^3cos3@ < 0
But the modulus is always positive, so r^3 > 0
Hence cos3@ < 0
And you don't even need general solutions.