Let z \:\varepsilon \:C be such that \left | z \right |<1. If \omega =\frac{5+3z}{5(1-z)} ,

then :

  • Option 1)

    5\:Re (\omega )>4

  • Option 2)

    5\:Im (\omega )>5

  • Option 3)

    5\:Re (\omega )>1

  • Option 4)

    5\:Im (\omega )<1

Answers (1)

\omega=\frac{5+3z}{5(1-z)}=\frac{5+3z}{5-5z}

\\5\:\omega-5\:\omega z= 5+3z\\\\\: 5\:\omega-5=3z+5\:\omega z

\\=(5\:\omega+3)z=5\:\omega-5\\\\\:z=\frac{5\:\omega-5}{5\:\omega+3}

\\\left | z \right |<1\\\\\:\Rightarrow\left | \frac{5\omega-5}{5\omega+3} \right |<1

\\ \left | 5\omega-5 \right |<\left | 5\omega+3 \right |\\\\\:\left | \omega -1 \right |<\left | \omega+3/5 \right |

Re(\omega )>1/5

 

 

 


Option 1)

5\:Re (\omega )>4

Option 2)

5\:Im (\omega )>5

Option 3)

5\:Re (\omega )>1

Option 4)

5\:Im (\omega )<1

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