Hi, could someone please help with this question?

We are given two complex numbers, after I sketched them on the argand diagram i did the following:

\(BC=2AB\) --------> \(OC-OB=2OB-2OA\) (Vectors) and solved for OC. But the answer is incorrect.

However, when using "2i" instead of "2", it gives the correct answer........ (Why??)

Also, another method is noticing when multiplying AB (v-u) by 2i, it gives z-v (I.e. rotation of AB onto BC by multiplying by i (as the angle is 90) and multiplying by 2 to give the magnitude) this method gives the correct answer. But, why not multiplication by -2i? (The rotation is really clockwise not anticlockwise?)

Also, if anyone has any other methods please share!

Thanks!

Guest May 6, 2022

#1**+2 **

You treated BC and AB as vectors. If \(\overrightarrow{BC} = 2\overrightarrow{AB}\), that means \(\overrightarrow{BC}\) is **parallel** to \(\overrightarrow{AB}\), which is not what you want, since \(\angle ABC = 90^\circ\).

Using 2i instead of 2 gives the vectors a rotation of 90 degrees, which gives the correct answer.

The multiplication by -2i gives a rotation of 90 degrees clockwise, maybe the resulting point C is not in the first quadrant, which isn't what the question wanted.

MaxWong May 6, 2022

#1**+2 **

Best Answer

You treated BC and AB as vectors. If \(\overrightarrow{BC} = 2\overrightarrow{AB}\), that means \(\overrightarrow{BC}\) is **parallel** to \(\overrightarrow{AB}\), which is not what you want, since \(\angle ABC = 90^\circ\).

Using 2i instead of 2 gives the vectors a rotation of 90 degrees, which gives the correct answer.

The multiplication by -2i gives a rotation of 90 degrees clockwise, maybe the resulting point C is not in the first quadrant, which isn't what the question wanted.

MaxWong May 6, 2022

#2**+2 **

Thank you!!

One last question: Is it ok to treat complex numbers as vectors? (Because I usually did and got correct answer, but here it is incorrect, as you said, because the vectors are not parallel. So, Is treating complex numbers as vectors only works if they are parallel, and if they are not parallel then we have to take into account the direction? I.e. rotation of the vector by angle say \(\theta\)?)

I mean is the following general formula correct (I am just guessing): \(BC=2AB \iff BC_{vector}=r(cos\theta+isin\theta)AB_{vector}\), so in this case, \(\theta=90\) and r = 2, and it worked. But, is this formula correct in general or? (Also the case when the two lines are parallel, if we used \(\theta=180\), then \(cos\theta=-1\), but we want it 1 to be correct I think? But if we used \(\theta=0\) then I think it works. So, is the angle between two parallel lines 180 or 0? (Because with 0, the direction is maintained but with 180, the direction switched, which won't give the correct answer I think).

Guest May 6, 2022

#3**+2 **

In general cases, you can treat complex numbers as vectors because there is a one-to-one correspondence \(\phi:\mathbb C \to \mathbb R^2\) defined by \(\phi (z) = \begin{bmatrix}\operatorname{Re}(z)\\\operatorname{Im}(z)\end{bmatrix}\).

You just have to be careful that if \(BC = 2AB\), that actually means \(\left\|\overrightarrow{BC}\right\| = 2 \left\|\overrightarrow{AB}\right\|\), not \(\overrightarrow{BC} =2 \overrightarrow{AB}\). This goes for any equation with lengths.

MaxWong
May 6, 2022