Question

No spams please. Please answer this fast.

Answer 1

Solution

(I)

• given=>z=2-3i
• now....

$\frac{1 - 2z}{z - 3} \\ = \frac{1 - 4 + 6i}{2 - 3i - 3} \\ = \frac{6i - 3}{ - 3i - 1} \\ = \frac{3 - 6i}{3i + 1} \\ = \frac{(3 - 6i)(3i - 1)}{(3i + 1)(3i - 1)} \\ = \frac{9i - 3 + 18 + 6i}{ - 9 - 1} \\ = \frac{15i + 15}{ - 10} \\ = - \frac{3}{2} + ( - \frac{3}{2} )i$

(ii)

$\frac{2 + i}{2 - 3i} = a + ib \\ = > \frac{(2 + i)(2 + 3i)}{(2 - 3i)(2 + 3i)} = a + ib \\ = > \frac{4 + 6i + 2i - 3}{4 + 9} = a + ib \\ = > \frac{1 + 8i}{13} = a + ib \\ = > \frac{1}{13} + \frac{8}{13} i = a +i b \\ compairing \: both \: sides \: we \: get...... \\ a = \frac{1}{13} \\ b = \frac{8}{13}$

Hope this help you........be happy....xd

Answer 2

$\huge\mathfrak{Hey\:mate}$

1 st case →

Given that z = 2 -3i

And $\huge\frac{1-2z}{z-3}$

Putting value of z here :-

$\frac{1 - 2(2 - 3i)}{2 - 3i - 3}$

$\frac{ - 3 + 6i}{ - 1 - 3i}$

Now multiplying and dividing by conjugate of-1-3i :-

$\frac{ - 3 + 6i}{ - 1 - 3i} \times ( \frac{ - 1 + 3i}{ - 1 + 3i} )$

$\frac{3 - 9i -6i - 18}{10 }$

$\frac{15 - 15i}{10}$

$\frac{3 - 3i}{2}$

$\frac{3}{2} - \frac{3i}{2} = a + ib$

2nd case →

Given that $\huge\frac{2+i}{2-3i}$ = A +iB

Now multiplying and dividing by conjugate of 2-3i :-

$\frac{2 + i}{2 - 3i} \times ( \frac{2 + 3i}{2 + 3i} )$

$\frac{1 + 8i}{13 } = a + ib$

By comparing both sides we got :-

$a = \frac{1}{13}$

$b = \frac{8}{13}$