Why is |z| = 1 a circle of radius one? (FP2)
So, basically |z| = 1 is equal to the set containing all complex numbers where their magnitude is equal to one. Also, by unraveling the definition of |z| we get that (x2+y2)1/2=1 which is the same as x2+y2=1 which we can identify as the circle with centred at (0,0) and radius 1.
CM
Related Maths A Level answers
All answers ▸Do the circles with equations x^2 -2x + y^2 - 2y=7 and x^2 -10x + y^2 -8y=-37 touch and if so, in what way (tangent to each other? two point of intersection?)
Find minimum and maximum of x^2+1 if they exist
The line l1 has equation y = −2x + 3. The line l2 is perpendicular to l1 and passes through the point (5, 6). (a) Find an equation for l2 in the form ax + by + c = 0, where a, b and c are integers.
Via the product rule, or otherwise, differentiate 'y = xsin(x)'.