Always appeal to geometry if you can. If you subtract i from a point z, you lower z by 1 unit (-i). And when you do, you end up on the 45 deg. line (PI/4). Therefore, z is all points on the 45 deg line RAISED by 1 unit, which line passes thru the points (-1+0i) & (0+i). In parametric form, this is (0+i) + t( (0+i)-(-1+0i) ) = i + ti + 1 = 1 + i(t+1), for t>0.
There is two cases Let the affixes M(Z), A(i) The expression becomes First case: Arg(Z-i)=(pi)/4+ k(pi),k is an entiger. The locus is a straight line passing by A(i) and it does not contains A(i) .with angle measure equal: (pi)/4+k(pi),k is entiger. Case two: Arg(Z-i)=(pi)/4+k(2pi),k entiger. The locus is half of straight line passing by A(i) and it does not contains A(i). with angle measure equal: (pi)/4+k(2pi),k is an entiger.
Always appeal to geometry if you can. If you subtract i from a point z, you lower z by 1 unit (-i). And when you do, you end up on the 45 deg. line (PI/4). Therefore, z is all points on the 45 deg line RAISED by 1 unit, which line passes thru the points (-1+0i) & (0+i).
In parametric form, this is (0+i) + t( (0+i)-(-1+0i) ) = i + ti + 1 = 1 + i(t+1), for t>0.
a+(a+1)i for a>0. -> y=x+1 for x>0.
Cool!
I think the theory of linear algebra is interesting, but performing the techniques needed to find answers is a lot of work without being interesting.
Arg z=arctan(y/x)
There is two cases
Let the affixes M(Z), A(i)
The expression becomes
First case:
Arg(Z-i)=(pi)/4+ k(pi),k is an entiger.
The locus is a straight line passing by A(i) and it does not contains A(i) .with angle measure equal:
(pi)/4+k(pi),k is entiger.
Case two:
Arg(Z-i)=(pi)/4+k(2pi),k entiger.
The locus is half of straight line passing by A(i) and it does not contains A(i).
with angle measure equal:
(pi)/4+k(2pi),k is an entiger.
note:arg(z)=arctan(b/a)
arctan((b-1)/a)=π/4
(b-1)/a=tan(π/4)=1
b-1=a
z = {x , x+1} where x > 0
by sybermath?
Yes
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Why don’t you like linear algebra?