Question

If \[{{Z}_{1}}=\left( 4,5 \right)\]and\[{{Z}_{2}}=\left( -3,2 \right)\], then \[\dfrac{{{Z}_{1}}}{{{Z}_{2}}}\]equals,
A) \[\left( \dfrac{-23}{12},\dfrac{-2}{13} \right)\]
B) \[\left( \dfrac{2}{13},\dfrac{-23}{13} \right)\]
C) \[\left( \dfrac{-2}{13},\dfrac{-23}{13} \right)\]
D) \[\left( \dfrac{-2}{13},\dfrac{23}{13} \right)\]

Answer 1

Hint: In the given question, we have been asked to simplify an expression. In order to simplify the given question, first we need to multiply the numerator and denominator by the conjugate of the denominator. Later we expand the numerator by using distributive property and then we simplify the expression after expanding. Then we will combine the like terms and simplify further. In this way, we will get our required solution.

Complete step-by-step solution:
We have given that,
\[{{Z}_{1}}=\left( 4,5 \right)\]And\[{{Z}_{2}}=\left( -3,2 \right)\]
Write the given coordinates in the complex form, we obtained
We have,
\[{{Z}_{1}}=\left( 4,5 \right)\]
\[{{Z}_{1}}=4+5i\]
And
\[{{Z}_{2}}=\left( -3,2 \right)\]
\[{{Z}_{2}}=-3+2i\]
Now,
\[\Rightarrow \dfrac{{{Z}_{1}}}{{{Z}_{2}}}=\dfrac{4+5i}{-3+2i}\]
Multiply the conjugate of denominator by numerator and denominator, we get
\[\Rightarrow \dfrac{{{Z}_{1}}}{{{Z}_{2}}}=\dfrac{4+5i}{-3+2i}\times \dfrac{-3-2i}{-3-2i}\]
Solving the above, we get
\[\Rightarrow \dfrac{{{Z}_{1}}}{{{Z}_{2}}}=\dfrac{4+5i\left( -3-2i \right)}{\left( -3+2i \right)\left( -3-2i \right)}\]
As we know that,
\[\Rightarrow \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]
Thus, applying this identity in the denominator, we get
\[\Rightarrow \dfrac{{{Z}_{1}}}{{{Z}_{2}}}=\dfrac{4+5i\left( -3-2i \right)}{{{\left( -3 \right)}^{2}}-{{\left( 2i \right)}^{2}}}\]
Solving the numerator using the distributive property of multiplication, we obtained
\[\Rightarrow \dfrac{{{Z}_{1}}}{{{Z}_{2}}}=\dfrac{-12-8i-15i-10{{i}^{2}}}{{{\left( -3 \right)}^{2}}-{{\left( 2i \right)}^{2}}}\]
Combining the like terms, we get
\[\Rightarrow \dfrac{{{Z}_{1}}}{{{Z}_{2}}}=\dfrac{-12-23i-10{{i}^{2}}}{{{\left( -3 \right)}^{2}}-{{\left( 2i \right)}^{2}}}\]
As we know that the value of an imaginary number \[{{i}^{2}}\]= -1,
Thus,
Putting the value of \[{{i}^{2}}\]= -1 in the above expression, we get
\[\Rightarrow \dfrac{{{Z}_{1}}}{{{Z}_{2}}}=\dfrac{-12-23i-10\left( -1 \right)}{{{\left( -3 \right)}^{2}}-4\left( -1 \right)}\]
Simplifying the above, we get
\[\Rightarrow \dfrac{{{Z}_{1}}}{{{Z}_{2}}}=\dfrac{-12-23i+10}{9+4}=\dfrac{-2-23i}{13}\]
Now,
\[\Rightarrow \dfrac{{{Z}_{1}}}{{{Z}_{2}}}=\dfrac{-2-23i}{13}=\left( \dfrac{-2}{13},\dfrac{-23}{13} \right)\]
Thus,
\[\Rightarrow \dfrac{{{Z}_{1}}}{{{Z}_{2}}}=\left( \dfrac{-2}{13},\dfrac{-23}{13} \right)\]

Hence, the option (C) is the correct answer.

Note: While solving these types of problems, students should remember that they need to factorize the given expression by rewritten the above expression in complex number form. Students need to be very careful while factorization, they need to write all the terms and the signs associated with it very explicitly and very carefully to avoid making errors. For factorization of the given expression while expanding these types of expression we will need to use the distributive property of multiplication. We need to multiply the numerator and denominator by the conjugate of the denominator so that we will get a real number in the denominator.