Question

(1+i)²/2+i find modulus and argument​

Answer 1

Answer:

modulus = 2/ $\sqrt{5}$ & argument --> α = tan^(-1) (-2)

Step-by-step explanation:

Here i^2 = ( -1 )

{(1+i)^2}/(2+i) = {1+(i^2)+2i}/(2+i) = (1 - 1 +2i)/(2+i) = 2i/(2+i)

now using rationalization we get

{2i x (2 - i)} / {(2=i) x (2-i)} = {4i -2(i^2)}/{4 - (i^2)} = (2+4i)/5

= 2/5 + 4i/5

now , modulus = $\sqrt{(2/5)^2 + (4/5)^2}$ = $\sqrt{(4/25) + (16/25}$ = $\sqrt{20/25}$ = $\sqrt{4/5}$

= 2/ $\sqrt{5}$

argument--> tanα = -(4/5)/(2/5) = -2 ==> α = tan^(-1) (-2)