I've been asked to prove that if | z | ≤ 1 , z ∈ C then | 3 z − i 3 + i z | ≤ 1. .I've tried letting z = a + b i and using the fact that | z | = a 2 + b 2 and expanding the expression within the modulus brackets but I'm having trouble showing that the remaining fraction is less than or equal to one.

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I&#039;ve been asked to prove that if       |    z    |    ≤  1  ,  z  ∈      C   then      |                  3        z        −        i                    3        +        i        z              |    ≤  1. .I&#039;ve tried letting   z  =  a  +  b  i and using the fact that       |    z      |    =            a              2              +          b              2             and expanding the expression within the modulus brackets but I&#039;m having trouble showing that the remaining fraction is less than or equal to one.

Anika Stevenson · Accepted Answer

Step 1Let   z  =  a  +  b  i . Then we will proove that:      |        3    z    −    i        |    ≤      |        3    +    i    z        |        9          a      2        +    (    3    b    −    1          )      2        ≤            a      2        +    (    3    −    b          )      2        9      a    2    +  9      b    2    −  6  b  +  1  ≤      a    2    +      b    2    +  9  −  6  b  8  (      a    2    +      b    2    )  ≤  8Which is true because       |    z      |    ≤  1

Ayanna Trujillo · Accepted Answer

Step 1To prove the inequality, since       |    3  z  −  i      |    ,      |    3  +  i  z      |    &amp;gt;  0  ,, it is enough to prove       |    3  z  −  i            |        2    ≤      |    3  +  i  z            |        2    . . Since z is complex, it can be written as   a  +  b  i for some   a  ,  b  ∈      R    . . Then, the equation becomes       |    3  a  +  (  3  b  −  1  )  i            |        2    ≤      |    (  3  −  b  )  +  a  i            |        2    . .By the definition of       |    ⋅      |    , , this is equivalent to   (  3  a      )    2    +  (  3  b  −  1      )    2    ≤  (  3  −  b      )    2    +      a    2    . . Further transforming, this becomes       a    2    +      b    2    ≤  1  , , or       |    a  +  b  i      |    ≤  1. .

I've been asked to prove that if | z | </mrow> ≤ 1

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