What does the complex number 1/2*(1+i) represent in linear mapping?

  • Thread starter Thread starter loy
  • Start date Start date
  • Tags Tags
    Linear Mapping
loy
Messages
16
Reaction score
0

Homework Statement


i recently saw a question about complex number, and its answer about the center of a circle is 1/2*(1+i). what does that mean?


Homework Equations


f:ℂ → ℝ^2


The Attempt at a Solution


since we define z=x+yi is an element of ℂ, so by the mapping above,
we can say that f:ℂ→ℝ^2 = f(x+yi)=(x,y)?
and by the inverse function of f,which is f-1(x,y)→(x+yi)?
From the statement above, f-1(1/2,1/2)=1/2+i/2 =1/2*(1+i) ?
 
Physics news on Phys.org
help me ~~~
 
Yes, that is the "complex plane" where we represent the complex number z= x+iy by the point (x, y). So the number 1/2(1+ i)= 1/2+ (1/2)i is represented by the point (1/2, 1/2).

A circle with center at (a, b) and radius r has equation (x- a)^2+ (y- b)^2= r^2 so a circle with center at (1/2, 1/2) has equation (x- 1/2)^2+ (y- 1/2)^2= r^2. You may also know that the "modulus" or "absolute value" of z= x+ iy is |z|= \sqrt{x^2+ y^2} so that circle can also be written as |z- (1/2)(1+ i)|= r.
 
Last edited by a moderator:
thanks very much
=)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top