How do you simplify \[{{\left( 5+6i \right)}^{2}}\]?
Answer
618.3k+ views
Hint: This question is from the topic of complex numbers. In solving this question, we will first remove the square and separate the term. After that, we will use the foil method and multiply the terms. After that, we will solve the further equation and get the answer. We will see an alternate method to solve this type of question.
Complete step by step solution:
Let us solve this question.
In this question, we have asked to simplify the term \[{{\left( 5+6i \right)}^{2}}\].
For simplifying the term \[{{\left( 5+6i \right)}^{2}}\], let us first remove the square and separate the term.
As we know that the square any term can also be written as the same term multiplied two times. Or, we can say that \[{{a}^{2}}\] can also be written as \[a\times a\]. So, we can write the term \[{{\left( 5+6i \right)}^{2}}\] as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=\left( 5+6i \right)\times \left( 5+6i \right)\]
Now, we will use the foil method here.
Formula for foil method is:
\[\left( a+b \right)\times \left( c+d \right)=ac+ad+bc+bd\]
So, using this foil method, we can write the above equation as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=5\times 5+5\times 6i+6i\times 5+6i\times 6i\]
The above equation can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=25+30i+30i+36{{i}^{2}}\]
As we know that the square of iota is negative of 1, so we can write the above equation as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=25+30i+30i+36{{i}^{2}}\]
The above equation can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=25+60i-36=-11+60i\]
Hence, we have solved the term \[{{\left( 5+6i \right)}^{2}}\]. The simplified value of \[{{\left( 5+6i \right)}^{2}}\] is \[-11+60i\].
Note: We should have a better knowledge in the topic of complex numbers. We can solve this question by an alternate method. For that method, we will use the formula: \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2\times a\times b\]
The term we have to solve is \[{{\left( 5+6i \right)}^{2}}\] can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}={{5}^{2}}+{{\left( 6i \right)}^{2}}+2\times 5\times 6i\]
The above equation can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=25+36{{i}^{2}}+60i\]
The above equation can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=25+36\left( -1 \right)+60i=25-36+60i\]
The above equation can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=-11+60i\]
Hence, we get the same answer from this method. So, we can use this method to solve this type of question.
Complete step by step solution:
Let us solve this question.
In this question, we have asked to simplify the term \[{{\left( 5+6i \right)}^{2}}\].
For simplifying the term \[{{\left( 5+6i \right)}^{2}}\], let us first remove the square and separate the term.
As we know that the square any term can also be written as the same term multiplied two times. Or, we can say that \[{{a}^{2}}\] can also be written as \[a\times a\]. So, we can write the term \[{{\left( 5+6i \right)}^{2}}\] as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=\left( 5+6i \right)\times \left( 5+6i \right)\]
Now, we will use the foil method here.
Formula for foil method is:
\[\left( a+b \right)\times \left( c+d \right)=ac+ad+bc+bd\]
So, using this foil method, we can write the above equation as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=5\times 5+5\times 6i+6i\times 5+6i\times 6i\]
The above equation can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=25+30i+30i+36{{i}^{2}}\]
As we know that the square of iota is negative of 1, so we can write the above equation as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=25+30i+30i+36{{i}^{2}}\]
The above equation can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=25+60i-36=-11+60i\]
Hence, we have solved the term \[{{\left( 5+6i \right)}^{2}}\]. The simplified value of \[{{\left( 5+6i \right)}^{2}}\] is \[-11+60i\].
Note: We should have a better knowledge in the topic of complex numbers. We can solve this question by an alternate method. For that method, we will use the formula: \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2\times a\times b\]
The term we have to solve is \[{{\left( 5+6i \right)}^{2}}\] can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}={{5}^{2}}+{{\left( 6i \right)}^{2}}+2\times 5\times 6i\]
The above equation can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=25+36{{i}^{2}}+60i\]
The above equation can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=25+36\left( -1 \right)+60i=25-36+60i\]
The above equation can also be written as
\[\Rightarrow {{\left( 5+6i \right)}^{2}}=-11+60i\]
Hence, we get the same answer from this method. So, we can use this method to solve this type of question.
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