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 $\dfrac{z}{w}$?
+532 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!!
+118608 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 \(\sqrt2\cdot i\)
