A) \[\frac{12}{5}+\frac{16}{5}\,i\]
B) \[\frac{4}{5}+\frac{6}{5}\,i\]
C) \[\frac{6}{5}-\frac{5}{2}\,i\]
D) None of these
Correct Answer: A
Solution :
Let \[z=\frac{12}{5}+\frac{16}{5}t\] \[\therefore \] \[\arg \,(z)=\frac{16}{12}=\frac{4}{3}>0\] and \[|z|=\sqrt{{{\left( \frac{12}{5} \right)}^{2}}+{{\left( \frac{16}{5} \right)}^{2}}}\] \[=\frac{1}{5}\sqrt{144+256}=\frac{29}{5}=4\] Now, \[|2-3i|=\sqrt{4+9}=\sqrt{13}\] \[\therefore \] \[|2-3i|<|z|\] Hence, option [a] is correct.
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