Complex numbers proof?

Well,

1) the result is obviously true for z=0

2) let z =/= 0

let' s use Euler's notation :

z = r e^(iθ) with r > 0 then we have the basic result : |z| = r

then :

z^n = r^n * [e^(iθ)]^n

= r^n * e^(inθ)

therefore :

|z^n| = |r^n * e^(inθ)|

= |r^n| * |e^(inθ)|

= |r^n| * 1

= r^n

= | z |^n

conclusion :

|z^n| = | z |^n

hope it' l lhelp !!

lemma

|xz| = |x||z|

z = a + bi

and x = c + di

xz = (ac - bd) + (ad + cb) I

|z| = (a^2+ b^2)^1/2

|x| = (c^2 + d^2)^1/2

|x||z| = (a^2+ b^2)^1/2(c^2 + d^2)^1/2

= (a^2c^2 + a^2 d^2 + b^2c^2 + b^2d^2)

|xz| = ((ac - bd)^2 + (ad + cb)^2)^1/2

= (a^2 c^2 - 2 abcd + b^2d^2 + a^2 d^2 + 2 abcd + c^2 b^2)^1/2

= (a^2 c^2 + b^2d^2 + a^2 d^2 + c^2 b^2)^1/2

▪

|z| = |z|

suppose

|z^n| = |z|^n

|z^n+1| = |z z^n | = |z||z|^n = |z|^n+1

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Take z=x+iy so |z|^n is √(x+iy)^n similarly |z^n| is √(x+iy)^n hence proved. Hope it helped