Question

If z1=(6+3i)and z2=(2-i) then z1/z2

Answer 1

$\huge\mathtt\color{orange} ANSWER :$

➪ z1/z2

➪6 + 3i / 2 - i

➪3

$<font color = green >$

➪3 is the answer .

Answer 2

Answer:

$\frac{1}{5}(9+12i)$

Step-by-step explanation:

Here $Z_1$ = 6+3i and $Z_2$ = 2-i

We do the operation $\frac{Z_1}{Z_2}$ as $z_{1} . \frac{1}{z_{2\\} }$

So, $\frac{1}{z_{2\\} }$ = $\frac{a}{a^{2}+b^{2} } + i \frac{-b}{a^{2} + b^{2} }$

         = $\frac{2}{2^{2}+1^{2} } +i \frac{-(-1)}{2^{2} +1^{2} }$      [Since a = 2 and b = -1]

         =   $\frac{2+i}{5}$    

$\frac{z_{1}}{z_{2} } = z_{1} . \frac{1}{z_{2} }$

      = $(6+3i)(\frac{2+i}{5} )$

    $=\frac{12}{5} + \frac{6i}{5} +\frac{6i}{5} + i^{2} \frac{3}{5} \\= \frac{12}{5} + \frac{6i}{5} +\frac{6i}{5} - \frac{3}{5} \\ = \frac{12}{5} - \frac{3}{5} + \frac{12i}{5} = \frac{9}{5} + \frac{12i}{5} \\= \frac{1}{5} (9 + 12i)$