How do you simplify\[\left( 2+3i \right)\left( 3-5i \right)\]?
Answer
618k+ views
Hint: In the given question, we have been asked to simplify the \[\left( 2+3i \right)\left( 3-5i \right)\]. In order to simplify the given expression, first we need to expand the given expression and simplify the numbers. Later we combine the like terms in the given expression and put the value of \[{{i}^{2}}\] in the given expression. And simplify the expression further and we will get our required solution.
Complete step-by-step solution:
We have given that,
\[\Rightarrow \left( 2+3i \right)\left( 3-5i \right)\]
Expanding the above expression, we get
\[\Rightarrow \left( 2+3i \right)\left( 3-5i \right)=\left( 2\times 3 \right)+\left( 2\times \left( -5i \right) \right)+\left( 3i\times 3 \right)+\left( 3i\times \left( -5i \right) \right)\]
Simplify the brackets in the above expansion, we get
\[\Rightarrow 6+\left( -10i \right)+9i+\left( -15{{i}^{2}} \right)\]
Combining the like terms in the above expression, we get
\[\Rightarrow 6-i+\left( -15{{i}^{2}} \right)\]
As, we know that
The value of \[{{i}^{2}}\]= -1.
Substituting it in the expression, we get
\[\Rightarrow 6-i+\left( -15\times \left( -1 \right) \right)\]
Simplifying the above, we get
\[\Rightarrow 6-i+15\]
\[\Rightarrow 21-i\]
Therefore,
\[\Rightarrow \left( 2+3i \right)\left( 3-5i \right)=21-i\]
Thus, by simplifying \[\left( 2+3i \right)\left( 3-5i \right)\] we get\[21-i\].
Hence, it is the required solution.
Note: Complex numbers are those numbers which can be written in the form of\[a+bi\], where ‘a’ is the real part and ‘b’ is the imaginary part of the given expression. An imaginary number can be said to be a complex number that can be written as a real number multiplied by the imaginary unit that is iota represented by \[i\]. While solving these types of question, sometimes question include the term \[{{i}^{2}}\] thus students need to remember the property of an imaginary unit that is \[{{i}^{2}}\] = -1. While solving these questions, the first is to expand the brackets i.e. open the brackets in the given expression and write the resultant expression i.e. complex number in the standard form.
Complete step-by-step solution:
We have given that,
\[\Rightarrow \left( 2+3i \right)\left( 3-5i \right)\]
Expanding the above expression, we get
\[\Rightarrow \left( 2+3i \right)\left( 3-5i \right)=\left( 2\times 3 \right)+\left( 2\times \left( -5i \right) \right)+\left( 3i\times 3 \right)+\left( 3i\times \left( -5i \right) \right)\]
Simplify the brackets in the above expansion, we get
\[\Rightarrow 6+\left( -10i \right)+9i+\left( -15{{i}^{2}} \right)\]
Combining the like terms in the above expression, we get
\[\Rightarrow 6-i+\left( -15{{i}^{2}} \right)\]
As, we know that
The value of \[{{i}^{2}}\]= -1.
Substituting it in the expression, we get
\[\Rightarrow 6-i+\left( -15\times \left( -1 \right) \right)\]
Simplifying the above, we get
\[\Rightarrow 6-i+15\]
\[\Rightarrow 21-i\]
Therefore,
\[\Rightarrow \left( 2+3i \right)\left( 3-5i \right)=21-i\]
Thus, by simplifying \[\left( 2+3i \right)\left( 3-5i \right)\] we get\[21-i\].
Hence, it is the required solution.
Note: Complex numbers are those numbers which can be written in the form of\[a+bi\], where ‘a’ is the real part and ‘b’ is the imaginary part of the given expression. An imaginary number can be said to be a complex number that can be written as a real number multiplied by the imaginary unit that is iota represented by \[i\]. While solving these types of question, sometimes question include the term \[{{i}^{2}}\] thus students need to remember the property of an imaginary unit that is \[{{i}^{2}}\] = -1. While solving these questions, the first is to expand the brackets i.e. open the brackets in the given expression and write the resultant expression i.e. complex number in the standard form.
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