2015/089) lf $z= \frac{a+3i}{2+ai}$ , a is real.

Show that there is only one value of a for which arg z = $\frac{pi}{4}$  and find it
Solution
$arg(z) = \frac{pi}{4}$

z must be located at first quadrant...


now $\frac{a+3i}{2+ai} = \frac{(a+3i)(2-ai)}{(4+a^2)}$

= $\dfrac{5a + (6-a^2)i}{4+ a^2}$

Arg z = 1 when $5a = 6- a^2$
or $a^2 + 5a – 6 = 0$
$(a+6)(a-1) = 0$
a = -6 or 1
a = -6 means $arg(z) = \frac{3pi}{4}$ and hence this is invalid value and so $a = 1$ is the correct ans.
Refer to https://in.answers.yahoo.com/question/index?qid=20111204094456AAQhinr