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

##### Member
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!!

umpbay

#### lolokay

##### Active Member
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

##### languages are my world <3
itd be interesting if any general maths student could help you on this, lol.

#### PC

##### Member
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

##### i am number -e^i*pi
Does that mean you have to be in Antarctica to answer the question?
oh my lol.

#### EvoRevolution

##### Member
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

##### Account Closed
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

##### Banned
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.