Question

if we say z1 / z2:
z1 - 2√2 ( cos(2π/3)+i sin2π/3))
z2- 2√3 ( cos (π/6)+i sin(π/6))
is the answer √6 / 3 or √6 /3 I?

Answer 1

If you mean:

\(z_1=2\sqrt{2}(\cos \frac{2\pi}{3}+i\sin\frac{2\pi}{3})\) and
\(z_2=2\sqrt{3}(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})\), then

\(\dfrac{z_1}{z_2}=\dfrac{2\sqrt{2}(\cos \frac{2\pi}{3}+i\sin\frac{2\pi}{3})}{2\sqrt{3}(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})}=\dfrac{\sqrt{6}(\cos \frac{2\pi}{3}+i\sin\frac{2\pi}{3})}{3(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})}\).

Now before doing anything algebraic with this number, remember what dividing two complex numbers means geometrically: you have to divide their magnitudes (absolute values, moduli, or whatever you call them...) and subtract their arguments.

We have already divided the moduli: \(|\dfrac{z_1}{z_2}|=\dfrac{\sqrt{6}}{3}\).
Now we have to subtract the arguments.

We see this:

\(\arg(z_1)=\frac{2\pi}{3}=\frac{4\pi}{6}\), \(\arg(z_2)=\frac{\pi}{6}\).

So all you have to do is: \(\arg\left( \dfrac{z_1}{z_2}\right)=\frac{4\pi}{6}-\frac{\pi}{6}=...\)

As you can see, the number you are looking for is very simple indeed!
Can you see now what it is?

Answer 2

the answer is then 3π / 6?

Answer 3

The argument of \(\dfrac{z_1}{z_2}\) is 3π/6 = π/2.
That puts is on the imaginary axis. (π/2 = 90 degrees).
Because we've found the magnitude to be \(\frac{\sqrt{6}}{3}\), the answer must be:

\(\frac{\sqrt{6}}{3}i\)

Answer 4

ok so it must have a I !! thanks

Answer 5

YW!