
If $z=x+iy$, where $x$ and $y$ belongs to $R$ and \[3x + \left( {3x - y} \right)i = 4 - 6i\] then $z =$
(A) \[\left( {\dfrac{4}{3}} \right) + i10\]
(B) \[\left( {\dfrac{4}{3}} \right) - i10\]
(C) \[ - \left( {\dfrac{4}{3}} \right) + i10\]
(D) \[ - \left( {\dfrac{4}{3}} \right) - i10\]
Answer
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Hint: Here, z is the combination of both the real number and imaginary number which is a complex number. So, we will equate the real part and then the imaginary part. We will get the value of x when we equate the real part and then the value of y when we equate the imaginary part. After that, substituting the values of x and y in the given z= x+iy, we will get the final value of z.
Complete step by step solution:
Given that, \[3x + \left( {3x - y} \right)i = 4 - 6i\]
Here, we will equate the real part and imaginary parts.
First, equate the real parts, we get,
\[\therefore 3x = 4\]
\[ \Rightarrow x = \dfrac{4}{3}\]
Next, equate the imaginary parts, we get,
\[\therefore 3x - y = - 6\]
Substituting the value of x in the above equation, we will get,
\[ \Rightarrow 3(\dfrac{4}{3}) - y = - 6\]
On evaluating the above equation, we will get,
\[ \Rightarrow 4 - y = - 6\]
By using the transposition method, keep only unknown term on LHS, we get,
\[ \Rightarrow - y = - 6 - 4\]
\[ \Rightarrow - y = - 10\]
\[ \Rightarrow y = 10\]
We are given that,
\[\therefore z = x + iy\] where x and y belong to R (real).
Substituting the value of x and y in the above equation, we will get,
\[ \Rightarrow z = \dfrac{4}{3} + i10\]
Rearranging this above equation, we will get,
\[ \Rightarrow z = \dfrac{4}{3} + 10i\]
Hence, for the given \[3x + \left( {3x - y} \right)i = 4 - 6i\] the value of \[z = \dfrac{4}{3} + 10i\].
Note: Complex numbers are the numbers that are expressed in the form of a+ib where a, b are real numbers and i is an imaginary number called iota. An imaginary number is usually represented by i or j, which is equal to \[\sqrt { - 1} \] . Thus, the square of an imaginary number is equal to a negative number (i.e. \[{i^2} = - 1\] ). The main application of these numbers is to represent periodic motions such as water waves, alternating current, light waves, etc.
Complete step by step solution:
Given that, \[3x + \left( {3x - y} \right)i = 4 - 6i\]
Here, we will equate the real part and imaginary parts.
First, equate the real parts, we get,
\[\therefore 3x = 4\]
\[ \Rightarrow x = \dfrac{4}{3}\]
Next, equate the imaginary parts, we get,
\[\therefore 3x - y = - 6\]
Substituting the value of x in the above equation, we will get,
\[ \Rightarrow 3(\dfrac{4}{3}) - y = - 6\]
On evaluating the above equation, we will get,
\[ \Rightarrow 4 - y = - 6\]
By using the transposition method, keep only unknown term on LHS, we get,
\[ \Rightarrow - y = - 6 - 4\]
\[ \Rightarrow - y = - 10\]
\[ \Rightarrow y = 10\]
We are given that,
\[\therefore z = x + iy\] where x and y belong to R (real).
Substituting the value of x and y in the above equation, we will get,
\[ \Rightarrow z = \dfrac{4}{3} + i10\]
Rearranging this above equation, we will get,
\[ \Rightarrow z = \dfrac{4}{3} + 10i\]
Hence, for the given \[3x + \left( {3x - y} \right)i = 4 - 6i\] the value of \[z = \dfrac{4}{3} + 10i\].
Note: Complex numbers are the numbers that are expressed in the form of a+ib where a, b are real numbers and i is an imaginary number called iota. An imaginary number is usually represented by i or j, which is equal to \[\sqrt { - 1} \] . Thus, the square of an imaginary number is equal to a negative number (i.e. \[{i^2} = - 1\] ). The main application of these numbers is to represent periodic motions such as water waves, alternating current, light waves, etc.
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