# Convert given two complex number division into polar form

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.

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

HINT

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

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

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.) $|q|= 1/4$