Question

prove that the modulus of complex number z=x+iy is √x²+y²​

Answer 1

If z = x+iy

|z| = √(x² + y²)

|z|² = x² + y² .…....….…... (1)

we know that,

(x + y)² = |x|² + ]y|² + 2|x||y| . .(2)

and (Ixl-lyl)² = |x|² + |y|² - 2|x|ly|

we know that, whole square is always greater than zero.

(Ixl-lyl)² ≥ 0

|x|2 + lyl2 - 2|x|ly| ≥ 0

|x|2 + lyl? > 2|x|ly|

from equation (2)

( \x[ + lyl )2 = |x|2 + lyl2 + |x|2 + ly|2

(x + y)² ≤ 2 (x² + y²)

(Ixl + lyl)² ≤ 2 |z|²both side square root

|x| + ly| ≤ √2 |z|

Hence proved

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Answer 2

Step-by-step explanation:

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