Question

find a and B if


I) 1/a-ib=3-2i









​

Answer 1

Explanation:

Answer: The required values of a and b are

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

13

3

, b=

13

2

.

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

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

a+ib

1

=3−2i (i)

We are to find the values of a and b.

From equation (i), we have

\begin{gathered}\dfrac{1}{a+ib}=3-2i\\\\\\\Rightarrow a+ib=\dfrac{1}{3-2i}.\end{gathered}

a+ib

1

=3−2i

⇒a+ib=

3−2i

1

.

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

\begin{gathered}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}.\end{gathered}

a+ib=

3−2i

1

⇒a+ib=

(3−2i)(3+2i)

3+2i

⇒a+ib=

3

2

−(2i)

2

3+2i

⇒a+ib=

9−4i

2

3+2i

⇒a+ib=

9+4

3+2i

[since i

2

=−1]

⇒a+ib=

13

3

+i

13

2

.

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

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

13

3

, b=

13

2

.

Thus, the required values of a and b are

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

13

3

, b=

13

2

.

Answer 2

Answer: The required values of a and b are

$a=\dfrac{3}{13},~~b=\dfrac{2}{13}.a= </p><p>13</p><p>3</p><p> </p><p> , b= </p><p>13</p><p>2</p><p> </p><p> .</p><p>$

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

$\dfrac{1}{a+ib}=3-2i~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i) </p><p>a+ib</p><p>1</p><p> </p><p> =3−2i (i)</p><p>$

We are to find the values of a and b.

From equation (i), we have

$\begin{gathered}\dfrac{1}{a+ib}=3-2i\\\\\\\Rightarrow a+ib=\dfrac{1}{3-2i}.\end{gathered} </p><p>a+ib</p><p>1</p><p> </p><p> =3−2i</p><p>⇒a+ib= </p><p>3−2i</p><p>1</p><p> </p><p> .</p><p>$

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

$\begin{gathered}a+ib=\dfrac{1}{3-2i}\\\\\\\Rightarrow a+ib=\dfrac{3+2i}{(3-2i)(3+2i)}\\\\\\\Rightarrow a+ib=\dfrac{3+2i}{3^2-(2i)^2}\\\\\\\Rightarrow a+ib=\dfrac{3+2i}{9-4i^2}\\\\\\\Rightarrow a+ib=\dfrac{3+2i}{9+4}~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }i^2=-1]\\\\\\\Rightarrow a+ib=\dfrac{3}{13}+i\dfrac{2}{13}.\end{gathered} </p><p>a+ib= </p><p>3−2i</p><p>1</p><p> </p><p> </p><p>⇒a+ib= </p><p>(3−2i)(3+2i)</p><p>3+2i</p><p> </p><p> </p><p>⇒a+ib= </p><p>3 </p><p>2</p><p> −(2i) </p><p>2</p><p> </p><p>3+2i</p><p> </p><p> </p><p>⇒a+ib= </p><p>9−4i </p><p>2</p><p> </p><p>3+2i</p><p> </p><p> </p><p>⇒a+ib= </p><p>9+4</p><p>3+2i</p><p> </p><p> [since i </p><p>2</p><p> =−1]</p><p>⇒a+ib= </p><p>13</p><p>3</p><p> </p><p> +i </p><p>13</p><p>2</p><p> </p><p> .</p><p>$

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

$a=\dfrac{3}{13},~~b=\dfrac{2}{13}.a= </p><p>13</p><p>3</p><p> </p><p> , b= </p><p>13</p><p>2</p><p> </p><p> .</p><p></p><p>Thus, the required values of a and b are</p><p></p><p>a=\dfrac{3}{13},~~b=\dfrac{2}{13}.a= </p><p>13</p><p>3</p><p> </p><p> , b= </p><p>13</p><p>2</p><p> </p><p> .$