Question

a) (3 + i)?
b)
6 + i
9.
If z and w are two complex numbers, prove that lz - w|>|z|-|w|​

Answer 1

Answer:

If z is given as x + iy, and w = u + iv, then |z| = √(x^2 + y^2), = r, say; and |w| = √(u^2 + v^2), = p, say Then we can also view z as r cis θ, and w as p cis φ (i.e., p (cos φ + i sin φ) in polar form; so that θ and φ can be determined from x, y, u, and v -- to within 2π.) Now if z + w = q, say, consider the triangle formed by z, w, and q. The angle opposite q is π - |φ - θ| . So we have |q|^2 = |z|^2 + |w|^2 + 2 |

Is this correct. Please somebody check it. And if there is any other method... then plaese do it.

z| |w| cos (θ + φ), which is < |z + w|^2, since cos x

Step-by-step explanation: