Question

(3+4i)^-2 Convert to modarg?

Answer 1

The complex number is  (3 + 4i)^-2 .

Convert the complex number 3 + 4i into polor form .

 r = (x² + y²)^1/2        θ = arc tan (y/x) 

 r = (3² + 4²)^1/2       θ = arc tan (4/3) 

 r = (9+16)^1/2          θ = arc tan (4/3) 

 r = (25)^1/2             θ = 53.130°

 r = √25 = 5           θ = 53.130°

The polar form of  x+iy is r(cos ø +i sin ø ) .

3 + 4i = 5 (cos 53.130° +i sin 53.130° ) 

De Movire's Theorem states that: 

(cos x + i sin x)^n = cos(nx) + i sin(nx) 

So  (3 + 4i)^-2 = (5)^-2 [cos(-2 *53.130°) + i sin (-2 * 53.130°)] 

(3 + 4i)^-2 = (1/25) [cos(-106.26°) + i sin (-106.26°)]  

So the modulus and argument form of complex number (3 + 4i)^-2 is 

(3 + 4i)^-2 = (1/25) [cos(-106.26°) + i sin (-106.26°)]  .