Express each of the following in the form (a + ib):

Solution:

Given: $\frac{3-4 i}{(4-2 i)(1+i)}$

Solving the denominator, we get

$\frac{3-4 i}{(4-2 i)(1+i)}=\frac{3-4 i}{4(1)+4(i)-2 i(1)-2 i(i)}$

$=\frac{3-4 i}{4+4 i-2 i-2 i^{2}}$

$=\frac{3-4 i}{4+2 i-2(-1)}$

$=\frac{3-4 i}{6+2 i}$

Now, we rationalize the above by multiplying and divide by the conjugate of 6 + 2i

$=\frac{3-4 i}{6+2 i} \times \frac{6-2 i}{6-2 i}$

$=\frac{(3-4 i)(6-2 i)}{(6+2 i)(6-2 i)}$

Now, we know that

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

So, eq. (i) become

$=\frac{(3-4 i)(6-2 i)}{(6)^{2}-(2 i)^{2}}$

$=\frac{3(6)+3(-2 i)+(-4 i)(6)+(-4 i)(-2 i)}{36-4 i^{2}}$

$=\frac{18-6 i-24 i+8 i^{2}}{36-4(-1)}\left[\because{\mathrm{i}}^{2}=-1\right]$

$=\frac{18-30 i+8(-1)}{36+4}$

$=\frac{18-30 i-8}{40}$

$=\frac{10-30 i}{40}$

$=\frac{10(1-3 i)}{40}$

$=\frac{1-3 i}{4}$

$=\frac{1}{4}-\frac{3}{4} i$