We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

Let $z$ and $w$ be the complex numbers in the picture below:

If $|z| = 4$ and $|w| = 2$, then what is the imaginary part of $\dfrac{z}{w}$?

Guest Jan 12, 2019

#1**0 **

ok just like the other one i did, the angle is 45 degrees. ignoring the diagram(again lol)

z=a+bi

w=c+di

a^2+b^2=16

c^2+d^2=4

if z=-4i then w is -2sqrt2-2isqrt2

divide w/z and take the reciprocal.

w/z = i(-2sqrt2-2isqrt2)/4 or (2sqrt2-2isqrt2)/4 which reduces to (sqrt2-isqrt2)/2.

to get z/w you have to take the reciprocal which is 2/(sqrt2-isqrt2). multiply by conjugate on top and bottom you get 2(sqrt2+isqrt2)/(2+2), which is

(sqrt2+isqrt2)/2 so the imaginary part is sqrt2/2.

HOPE THIS HELPED!!

asdf335 Jan 12, 2019

#3**+2 **

If $|z| = 4$ and $|w| = 2$, then what is the imaginary part of $\dfrac{z}{w}$?

\(w=2e^{i \theta}\\ z=4e^{ i(\theta+\frac{\pi}{4} )}\\~\\ \frac{z}{w}=\frac{4e^{ i(\theta+\frac{\pi}{4} )}}{2e^{i \theta}}\\ =2e^{i\theta+\frac{\pi i}{4}-i\theta}\\ =2e^{\frac{\pi i}{4}}\\ =2cos\frac{\pi}{4}+(2sin\frac{\pi}{4})i\\ =\frac{2}{\sqrt2}+\frac{2}{\sqrt2}i\\ =\sqrt2+\sqrt2i\)

So the imaginary part is \(\sqrt2\cdot i\)

.Melody Jan 12, 2019