0
$\begingroup$

Sorry if I am breaking any rule. But I really need help with polar form because I have an exam Tomorrow.

Suppose $z=1+i$ and $w=1−i\sqrt3$. Write $q=z^6/w^5$ in polar form and calculate its modulus.

What I have tried so far: First, I found z

z=$√2(\cos \Pi/4+ i sin \Pi/4)$

then W

r = $√(1+3) = 2$ z=$√2(\cos\theta + i sin\theta)$

but I don't know how to find theta here.

Thanks in advance!

$\endgroup$
5
  • 1
    $\begingroup$ You, don't just ask with urgency. Provide context with urgency! $\endgroup$
    – Parcly Taxel
    Jan 28, 2018 at 14:51
  • $\begingroup$ Please edit the question to show us how you have started. Then perphaps we can help. Can you write $1+i$ in polar form? $\endgroup$
    – Ethan Bolker
    Jan 28, 2018 at 14:51
  • $\begingroup$ What is $\theta$ in your formula for $z$? Then do $w$. $\endgroup$
    – Ethan Bolker
    Jan 28, 2018 at 14:58
  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$
    – user
    Jan 31, 2018 at 22:29
  • $\begingroup$ what is it math.stackexchange.com/questions/2625017/…? $\endgroup$
    – Phoenix404
    Feb 12, 2018 at 14:30

2 Answers 2

Reset to default
0
$\begingroup$

HINT

Write z and w in exponential form then compute $z^6$ and $w^5$ then divide.

$\endgroup$
5
  • $\begingroup$ I did polar form for z but stuck on w, don't know how to find theta $\endgroup$
    – Phoenix404
    Jan 28, 2018 at 14:58
  • $\begingroup$ Draw it and it will be extremely obvious. $\endgroup$
    – orion
    Jan 28, 2018 at 15:02
  • $\begingroup$ Sorry I mean exponential, for w $\theta=-\pi/3$ $\endgroup$
    – user
    Jan 28, 2018 at 15:02
  • $\begingroup$ could you calculate w only? $\endgroup$
    – Phoenix404
    Jan 28, 2018 at 15:22
  • $\begingroup$ w has modulus 2 thus $$w =2\cdot e^{-i\frac{\pi}{3}}$$ $\endgroup$
    – user
    Jan 28, 2018 at 15:26
0
$\begingroup$

Asked : $z=1+iz=1+i$

$w=1−i3$

$q=z^6/w^5$

1.) Calculate the exponential form of q

2.) Calculate the modulus

1.)

$z = √2(cos(\pi/4)+ isin(\pi/4), z= \sqrt2e\^(i\pi/4)$ $w = √2(cos(-\pi/3)+ isin(-\pi/3), w= \sqrt2e\^(-i\pi/4)$

$z^6 = 8e\^(i3/2\pi)$

$w^5=32e\^(-i5/3\pi9$

$q= 1/4e\^(i19/6\pi)$

2.) $|q|= 1/4$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .