Question

1/a+ib =3-2i, find a & b​

Answer 1

Answer:  The required values of a and b are

$a=\dfrac{3}{13},~~b=\dfrac{2}{13}.$

Step-by-step explanation:  We are given the following :

$\dfrac{1}{a+ib}=3-2i~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We are to find the values of a and b.

From equation (i), we have

$\dfrac{1}{a+ib}=3-2i\\\\\\\Rightarrow a+ib=\dfrac{1}{3-2i}.$

To rationalize the denominator on the right-hand side of the above equation, we need to multiply both the numerator and denominator by the conjugate of (3 - 2i), that is, (3 + 2i).

So, we get

$a+ib=\dfrac{1}{3-2i}\\\\\\\Rightarrow a+ib=\dfrac{3+2i}{(3-2i)(3+2i)}\\\\\\\Rightarrow a+ib=\dfrac{3+2i}{3^2-(2i)^2}\\\\\\\Rightarrow a+ib=\dfrac{3+2i}{9-4i^2}\\\\\\\Rightarrow a+ib=\dfrac{3+2i}{9+4}~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }i^2=-1]\\\\\\\Rightarrow a+ib=\dfrac{3}{13}+i\dfrac{2}{13}.$

Equating the real and imaginary parts of both sides in the above equation, we get

$a=\dfrac{3}{13},~~b=\dfrac{2}{13}.$

Thus, the required values of a and b are

$a=\dfrac{3}{13},~~b=\dfrac{2}{13}.$

Answer 2

Answer:a=3/13 ,b=2/13

Step-by-step explanation:1/a+ib=3-2i

a+ib=1/3-2i (By Rationalizing The Denomonator)

a+ib=1 (3+2i)/(3-2i) (3+2i)

a+ib=3+2i/(3×3)-(2i×2i)

a+ib=3+2i/9-4i*i (i*2=-1)

a+ib=3+2i/9-(4×-1)

a+ib=3+2i/9-(-4)

a+ib=3+2i/9+4

a+ib=3+2i/13

Therefore a=3/13 , b=2/13