polar form, urgent

**Revolt** · January 13th 2008, 09:03 PM

ok, exam in 2 hours and i dont need the answer to this question, i need a list of the steps i take in order to do it, would be hugely grateful.

z = 1 - i
w = 1+ i

a) write in polar form
b) evaluate z^3/w^7 in form a+ib
c) find all cube roots of z leaving answers in polar form

thanks if anyone can help

**Jhevon** · January 13th 2008, 09:26 PM

don't you think this is cutting it a bit close. asking for help 2 hours before an exam?

Originally Posted by Revolt

ok, exam in 2 hours and i dont need the answer to this question, i need a list of the steps i take in order to do it, would be hugely grateful.

z = 1 - i
w = 1+ i

a) write in polar form

the polar form of a complex number $a + ib$ is: $re^{i \theta}$

where $r = \sqrt{a^2 + b^2}$ and $\theta = \arctan \frac ba$ (usually)

b) evaluate z^3/w^7 in form a+ib

find and simplify $\frac {z^3}{w^7}$ with $z$ and $w$ in their polar forms. then convert back to $a + ib$ form

c) find all cube roots of z leaving answers in polar form

see post #4 here

of course, we would write $e^{i \theta}$ instead of $\cos \theta + i \sin \theta$ as was done there

try them and tell me what you get. it will be good practice for that exam of yours

**CaptainBlack** · January 13th 2008, 09:31 PM

Originally Posted by Revolt

ok, exam in 2 hours and i dont need the answer to this question, i need a list of the steps i take in order to do it, would be hugely grateful.

z = 1 - i
w = 1+ i

a) write in polar form
b) evaluate z^3/w^7 in form a+ib
c) find all cube roots of z leaving answers in polar form

thanks if anyone can help

(a)

$z=r e^{i \theta}=r \cos(\theta) + r \sin(\theta) = 1-i$

so:

$r=\sqrt{2}$ ,

and $\theta$ is the angle such that:

$\cos(\theta)=1/\sqrt(2)$

and

$\sin(\theta)=-1/\sqrt(2).$

A similar procedure will turn w into polar form: $w=s e^{i \phi}$

(b) $z^3/w^7 = (r^4/s^7)(e^{i 3 \theta}/e^{i 7 \phi})= (r^4/s^7)e^{i (3 \theta-7 \phi)}$

which you should be able to turn back into cartesian form.

(c) $z=r e^{i \theta}$ where $r$ and $\theta$ take the values found earlier.

Then we may write:

$z=r e^{i \theta+2 i n \pi},\ n\in \mathbb{Z}$ (an integer)

then

$z^{1/3}=r^{1/3} e^{(i \theta+2 i n \pi)/3},\ n\in \mathbb{Z}$

and there are just three distinct values for this corresponding to $n=0,1,2$

RonL

**noreaction** · January 13th 2008, 09:39 PM

z = 1 - i
w = 1+ i

a) write in polar form

Convert to polar form:

find the modulus $|z|$ (also written as r), which is $\sqrt{(x^2 + y^2)}$
in the case of z $\sqrt{(1^2 + 1^2)}= \sqrt 2$
z = r = $\sqrt2$
Then find the argument $\theta$ . Since $tan \theta = \frac {rise}{run}$ you can easily find the argument here.

Polar form would be $r (cos \theta + isin \theta)$

b) evaluate z^3/w^7 in form a+ib

To find exponents of complex numbers, use De Moivre's theorem

If $z = r (cos \theta + isin \theta)$
then $z^n = r^n (cos (n\theta) + isin (n\theta))$

After this, convert back to a+ib form by multiplying r to $(cos \theta + isin \theta)$

c) find all cube roots of z leaving answers in polar form

Assume $z = m^3$ where m is a cube root of z

so $m = \sqrt[3]r (cos \frac {\theta}{3} + i sin \frac {\theta}{3})$

**CaptainBlack** · January 13th 2008, 09:42 PM

Originally Posted by Jhevon

where $r = \sqrt{a^2 + b^2}$ and $\theta = \arctan \frac ba$ (usually)

$\arctan$ does not do here as we need the angle in the range $[0 , 2 \pi)$ (or equivalent) but $\arctan$ gives an answer in the range $(\pi/2,\pi/2]$ (or equivalent).

RonL

**CaptainBlack** · January 13th 2008, 09:44 PM

Originally Posted by noreaction

c) find all cube roots of z leaving answers in polar form

Assume $z = m^3$ where m is a cube root of z

so $m = \sqrt[3]r (cos \frac {\theta}{3} + i sin \frac {\theta}{3})$

How many cube roots of $z$ does that give us? Should there not be 3?

RonL

**noreaction** · January 13th 2008, 09:47 PM

Sorry, i forgot to put $+2k\pi$ for the angles

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