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

Question:

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

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

Solution:

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

We solve the above equation by using the formula

$(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$

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

$=\frac{1+8 i^{3}+6 i+12 i^{2}}{2-i+2 i-i^{2}}$

$=\frac{1+8 i \times i^{2}+6 i+12(-1)}{2+i-(-1)}\left[\because i^{2}=-1\right]$

$=\frac{1+8 i(-1)+6 i-12}{2+i+1}$

$=\frac{1-8 i+6 i-12}{3+i}$

$=\frac{-11-2 i}{3+i}$

Now, we rationalize the above by multiplying and divide by the conjugate of 3 + i

$=\frac{-11-2 i}{3+i} \times \frac{3-i}{3-i}$

$=\frac{(-11-2 i)(3-i)}{(3+i)(3-i)} \ldots$ (i)

Now, we know that,

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

So, eq. (i) become

$=\frac{(-11-2 i)(3-i)}{(3)^{2}-(i)^{2}}$

$=\frac{-11(3)+(-11)(-i)+(-2 i)(3)+(-2 i)(-i)}{9-i^{2}}$

$=\frac{-33+11 i-6 i+2 i^{2}}{9-(-1)}\left[\because i^{2}=-1\right]$

$=\frac{-33+5 i+2(-1)}{9+1}\left[\because i^{2}=-1\right]$

$=\frac{-33+5 i-2}{10}$

$=\frac{-35+5 i}{10}$

$=\frac{5(-7+i)}{10}$

$=\frac{-7+i}{2}$

$=\frac{-7}{2}+\frac{1}{2} i$