complex no. qs (1 Viewer)

[Evergreen]

Hey ppl im having trouble with some questions which may seem pretty easy to you but i only started this topic yesterday. anyway;

Q1. A point A in an argand diagram represents the complex number 1+i. Find the complex number represented by B if OBA is an equalateral triangle.

Q2. OABC is a rhombus in an argand diagram, where O is the origin. The point A is (1,2), If angle BOA=30 degrees, and B is in the 2nd quadrant, find the complex numbers representing B and C

Q3. OPQR is a square in an argand diagram, where O is the origin. The point P represents the complex number z given by z=r(cos 30+isin30). Find the complex numbers represented by Q and R.

Please show full working out. Also i don't mind if you just do 1 question because i usually understand new types of questions if given an example. thanks

 

[Evergreen]

dw about Q3... just solved it using just using argand diagram...

still cant figure 1 and 2 though.......

 

[roadrage75]

for question 1,

line OB is simply line OA rotated anticlockwise by 60 degrees (modOB = modOA because it is an equallateral triagnle).

hence OB = OAcis(pi/3) <-- ((note zcisA = z rotated A radians anticlockwise)

hence since OA = 1+ i , OB = (1+i)(cis (pi/3)) = (1+i)(1/2 + i√3/2)
= (1-√ 3)/2 + i(1+√ 3)/2


NOTE ON MY DIAGRAM, the clockwise order of points are O, B, A

hopefully you will be able to do question 2 now.

 

[ngogiathuan]

For Q1, Let O represents complex number 0, A repre. a and B repre. b.Rotate vector OA 60 degrees anticlockwise or clockwise = vector OB
-> (b-0)=(a-0) cis (+-60 degrees)

For Q2, Let O, A, C, B repre. o,a,c,b respectively.
Rotate vector OA 30 degrees ANTICLOCKWISE=vector OB
-> (b-0)=(a-0) cis (30)
vector OA=vector BC therefore
(a-0)=(c-b)
Or you can rotate vector BO 150 degrees anticlockwise to get C doesn't matter