Solveeit Logo
Login

Question

Question: What is the value of \[{{(1+i)}^{5}}+{{(1-i)}^{5}}\] where \[i=\sqrt{-1}\] ? 1). \[-8\] 2). \[8\...

What is the value of (1+i)5+(1−i)5{{(1+i)}^{5}}+{{(1-i)}^{5}} where i=−1i=\sqrt{-1} ?
1). −8-8
2). 88
3). 8i8i
4). −8i-8i

Explanation

Solution

To solve this type of problem, first try to break the powers and try to simplify the equation using some formulas and then try to simplify it more using the powers of imaginary numbers and then add both of them and after that you will get your required answer.

Complete step-by-step solution:
Imaginary numbers can be defined as the numbers when squared gives a negative number as a result. In other words, imaginary numbers are defined as the square root of negative numbers when they are of an indefinite value. It is mostly real written in the form of real numbers multiplied by a guessing unit called as “i”.
We can also apply operations on the imaginary numbers such as all the basic arithmetic operations: addition, subtraction, multiplication and division.
Now, let’s understand the complex number.
Complex Number can be defined as the combination of both a real number and an imaginary number. Complex number also follows the commutative law and associative law of addition and multiplication and also obeys the distributive law. There is a property of complex numbers that when we do the addition of two complex numbers it will always result in a real number.
Real numbers can be defined as any number that is on the number system such as positive, negative, zero, integer, rational, irrational, dfractions etc. The absolute value of a real number is the given number itself.
Now, according to the question:
It is given that: i=−1i=\sqrt{-1}
Squaring on both sides:
⇒i2=(−1)2\Rightarrow {{i}^{2}}={{(\sqrt{-1})}^{2}}
⇒i2=−1\Rightarrow {{i}^{2}}=-1
As given in the question:
(1+i)5+(1−i)5{{(1+i)}^{5}}+{{(1-i)}^{5}} ............(1)............(1)
Now, first solving (1+i)5{{(1+i)}^{5}} by breaking the powers, such as:
⇒(1−i)3.(1−i)2\Rightarrow {{(1-i)}^{3}}.{{(1-i)}^{2}}
Now, using formula: (a+b)3=a3+b3+3a2b+3ab2{{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}} and (a+b)2=a2+b2+2ab{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab
⇒(1+i3+3i+3i2).(1+i2+2i)\Rightarrow (1+{{i}^{3}}+3i+3{{i}^{2}}).(1+{{i}^{2}}+2i)
Substituting values of i3=−i{{i}^{3}}=-i and i2=−1{{i}^{2}}=-1 , we will get as:
=(1−i+3i−3).(1−1+2i)= (1-i+3i-3).(1-1+2i)
=(1−i+3i−3).(2i)= (1-i+3i-3).(2i)
=(2i−2).(2i)= (2i-2).(2i)
=(4i2−4i)= (4{{i}^{2}}-4i)
=(−4−4i)= (-4-4i)
=(1+i)5=(−4−4i)= {{(1+i)}^{5}}=(-4-4i) ............(2)............(2)
Now, solving (1−i)5{{(1-i)}^{5}} by breaking the powers, such as:
⇒(1−i)3.(1−i)2\Rightarrow {{(1-i)}^{3}}.{{(1-i)}^{2}}
Now, using formula: (a−b)3=a3−b3−3a2b+3ab2{{(a-b)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}} and (a−b)2=a2+b2−2ab{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab
=(1−i3−3i+3i2).(1+i2−2i)= (1-{{i}^{3}}-3i+3{{i}^{2}}).(1+{{i}^{2}}-2i)
Substituting values of i3=−i{{i}^{3}}=-i and i2=−1{{i}^{2}}=-1 , we will get as:
=(1+i−3i−3).(1−1−2i)= (1+i-3i-3).(1-1-2i)
=(1+i−3i−3).(−2i)= (1+i-3i-3).(-2i)
=(−2i−2).(−2i)= (-2i-2).(-2i)
=(4i2+4i)= (4{{i}^{2}}+4i)
=(4i−4)= (4i-4)
=(1−i)5=(4i−4)= {{(1-i)}^{5}}=(4i-4) ............(3)............(3)
Now, substituting values from equation (2)(2) and equation (3)(3) in the equation (1)(1)
=(1+i)5+(1−i)5= {{(1+i)}^{5}}+{{(1-i)}^{5}}
=(−4−4i)+(4i−4)= (-4-4i)+(4i-4)
=−4−4i+4i−4= -4-4i+4i-4
∴−8\therefore -8
Hence, the correct option from all above option is 11.

Note: There are different properties of a complex number such as: If we are given two complex numbers and the sum of both of them is real and the product of both of them is also real, then both of these complex numbers are conjugate to each other. In other words, multiplication of two conjugate complex numbers will always result in a real number.