Can you give a better solution? (1 Viewer)

[_luke_]

So I've got this complex numbers question from the 1991 HSC Paper. I know its an easy question, but was wondering if my answer could become any simpler?

i) is easy
ii) i said put multiply E by i (rotation anti-clockwise 90*), no multiplication by factor c due to square having the same length.
iii) just add them together (w1 + the value of e).

Like I said, it seems simple but wondering if there are any simpler combinations that I haven't thought of? Give me your opinion.

Cheers

 

[leehuan]

Multiplication by i cannot be made easier for part ii either.

Am I going crazy... or isn't D the value of A + the value of F, not E?

But that seems to be the way to go about it

 

[_luke_]

Yeah I was thinking F, but wrote down E, my bad sorry. Addition by E would've formed the point B (w2).

 

[dan964]

Yes, the solutions have...
===
1991 HSC Q2d
(i)
OE || AB and OE = AB

:. OABE is a parallelogram.
Let E correspond to the complex number z, then, since addition of complex numbers corresponds to a
parallelogram construction in the complex plane,




i.e. E corresponds to the complex number

(ii) OF perpendicular to OE and OF = OE

Since multiplying a complex number by i corresponds to an anticlockwise rotation about the origin through 90°,

F corresponds to the complex number .

(iii) Since AD || OF and AD = OF,
D corresponds to the complex number