Question

What is $\dfrac{{{(1+i)}^{4n+5}}}{{{(1-i)}^{4n+3}}}$equal to, where n is a natural number and $i=\sqrt{-1}$?
A). $2$
B). $2i$
C). $-2i$
D). $i$

Answer 1

In this problem, the question is related to the concept of complex numbers. For this particular problem expression is given. We have to conjugate that expression that means multiply the numerator and denominator with ${{(1+i)}^{4n+3}}$and after simplification and using the value of $i=\sqrt{-1}$we get the value of expression.

Complete step-by-step solution:
In this type of problem, expression is given that is $\dfrac{{{(1+i)}^{4n+5}}}{{{(1-i)}^{4n+3}}}$
To solve this expression first of all we need to multiply by its conjugate of its denominator. That means in denominator ${{(1-i)}^{4n+3}}$we have to take conjugate of this ${{(1+i)}^{4n+3}}$then we have to multiply ${{(1+i)}^{4n+3}}$on numerator as well as denominator.
$\Rightarrow \dfrac{{{(1+i)}^{4n+5}}}{{{(1-i)}^{4n+3}}}\times \dfrac{{{(1+i)}^{4n+3}}}{{{(1+i)}^{4n+3}}}$
By simplifying further we get:
$\Rightarrow \dfrac{{{(1+i)}^{4n+5}}\times {{(1+i)}^{4n+3}}}{{{(1-i)}^{4n+3}}\times {{(1+i)}^{4n+3}}}$
Further solving this by using the property of indices ${{(a)}^{m}}\times {{(a)}^{n}}={{a}^{m+n}}$we get:
$\Rightarrow \dfrac{{{(1+i)}^{4n+5+}}^{4n+3}}{{{(1-i)}^{4n+3}}\times {{(1+i)}^{4n+3}}}$
After simplifying this we get:
$\Rightarrow \dfrac{{{(1+i)}^{4n+5+}}^{4n+3}}{{{\left[ (1-i)\times (1+i) \right]}^{4n+3}}}$
If you see the above equation then you can apply the property of $(a-b)(a+b)={{a}^{2}}-{{b}^{2}}$.
$\Rightarrow \dfrac{{{(1+i)}^{8n+8}}}{{{\left( {{1}^{2}}-{{i}^{2}} \right)}^{4n+3}}}$
After simplifying this we get:
$\Rightarrow \dfrac{{{\left[ {{(1+i)}^{2}} \right]}^{4n+4}}}{{{\left( {{1}^{2}}-{{i}^{2}} \right)}^{4n+3}}}$
You can notice in the above equation that you can apply the property of ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$in the above equation we get:
$\Rightarrow \dfrac{{{\left[ 1+2i+{{i}^{2}} \right]}^{4n+4}}}{{{\left( {{1}^{2}}-{{i}^{2}} \right)}^{4n+3}}}$
As we know that the value of $i=\sqrt{-1}$therefore, ${{i}^{2}}=-1$substitute this value in above equation we get:
$\Rightarrow \dfrac{{{\left[ 1+2i-1 \right]}^{4n+4}}}{{{\left( 1-(-1) \right)}^{4n+3}}}$
After simplifying this we get:
$\Rightarrow \dfrac{{{\left[ 2i \right]}^{4n+4}}}{{{\left( 2 \right)}^{4n+3}}}$
Bu simply splitting the term in numerator and denominator we get:
$\Rightarrow \dfrac{{{\left( 2 \right)}^{4n}}\times {{\left( 2 \right)}^{4}}\times {{\left( i \right)}^{4n+4}}}{{{\left( 2 \right)}^{4n}}\times {{\left( 2 \right)}^{3}}}$
By simplifying this we get:
$\Rightarrow \dfrac{{{\left( 2 \right)}^{4n}}\times {{\left( 2 \right)}^{4}}\times {{\left[ {{\left( i \right)}^{4}} \right]}^{n+1}}}{{{\left( 2 \right)}^{4n}}\times {{\left( 2 \right)}^{3}}}$
As we know that ${{i}^{4}}=1$substitute this value in above equation
And also by further solving this equation we get:
$\Rightarrow 2\times 1=2$
Therefore, $\dfrac{{{(1+i)}^{4n+5}}}{{{(1-i)}^{4n+3}}}=2$
So, the correct option is “option A”.

Note: While doing the simplification of a fraction related to complex numbers then keep in mind that we need to multiply the numerator and denominator by its conjugate of denominator. Conjugate means all you have to do is change the sign between two terms in the denominator.