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How do you simplify \[\dfrac{5}{1+i}\]?

Answer
VerifiedVerified
426.6k+ views
Hint: The given expression is of the form \[\dfrac{a+bi}{c+di}\], but as the imaginary part of the numerator is zero for this, we can also express it as \[\dfrac{a}{c+di}\]. To simplify these types of expression means we need to get rid of the imaginary number in the denominator. To do this, we need to multiply and divide the given expression with the conjugate of the imaginary number in the denominator. The conjugate of the number \[c+di\] is \[c-di\]. Here, \[i\] is an imaginary number equals \[\sqrt{-1}\].

Complete step by step solution:
We are asked to simplify \[\dfrac{5}{1+i}\]. This expression is of the form of \[\dfrac{a}{c+di}\], here the values of \[a,c,d\] as \[5,1,1\] respectively. As we know that to simplify these types of expressions, we need to multiply it by the conjugate of the complex number in the denominator. The complex number in the denominator is \[1+i\], so its complex conjugate will be \[1-i\]. By multiplying and dividing the given expression by this conjugate, we get
\[\Rightarrow \dfrac{5}{1+i}\times \dfrac{1-i}{1-i}\]
First, let’s simplify the numerator \[5\left( 1-i \right)\]. To simplify it, we need to expand the bracket using the distributive property which states that \[c\left( a+b \right)=ca+cb\]. Thus, the expansion of \[5\left( 1-i \right)\] is done as,
\[\begin{align}
  & \Rightarrow 5\left( 1-i \right) \\
 & \Rightarrow 5-5i \\
\end{align}\]
The denominator is \[\left( 1+2i \right)\left( 1-2i \right)\]. Its expansion will be
\[\begin{align}
  & \Rightarrow \left( 1+i \right)\left( 1-i \right) \\
 & \Rightarrow {{\left( 1 \right)}^{2}}-{{\left( i \right)}^{2}} \\
 & \Rightarrow 1-{{\left( \sqrt{-1} \right)}^{2}}=1-(-1) \\
 & \Rightarrow 2 \\
\end{align}\]
 \[\Rightarrow \dfrac{5}{1+i}\times \dfrac{1-i}{1-i}\]
Using the expansions of the numerator and the denominator, we can simplify the expression as
\[\Rightarrow \dfrac{5}{1+i}\times \dfrac{1-i}{1-i}\]
\[\begin{align}
  & \Rightarrow \dfrac{\left( 5 \right)}{\left( 1+i \right)}\dfrac{\left( 1-i \right)}{\left( 1-i \right)} \\
 & \Rightarrow \dfrac{5-5i}{2} \\
\end{align}\]
Separating the denominator, we get
\[\Rightarrow \dfrac{5}{2}-\dfrac{5i}{2}\]
This is the simplified form of the given expression.

Note:
The general expression for these types of questions is \[\dfrac{a+bi}{c+di}\]. To simplify these types of expressions also, we need to multiply it by the conjugate of the denominator which is \[c-di\]. But, as we have an imaginary number in the numerator also, we need to simplify it as well.