Question

express (1-i^3/1+i^3) ^2 in the form of a+ib​

Answer 1

Answer:

step by step explanation:

given expression (1-i^3/1+i^3)^2

first we solve (1+i^3/1-i^3)

by using (a+b)^3=a^3+b^3+3ab(a+b) and

(a-b)^3=a^3-b^3-3ab(a-b)

(1+i^3)=1+i^3+3i(1+i)=1+i^2*i+3i+3i^2=1+(-1)i+3i+3(-1)

=1-i+3i-3=-2+2i=-2(1-i) [here i^2=-1]

similarly

(1-i^3)=-2(1+i)

now (1-i^3/1+i^3)^2=(-2(1+i)/-2(1-i))^2

=(1+i/1-i)^2

by using (a+b)^2=a^2+b^2+2ab

(a-b)^2=a^2+b^2-2ab

(1+i/1-i)^2=1+i^2+2i/1+i^2-2i

=1+(-1)+2i/1+(-1)-2i

=1-1+2i/1-1-2i

=2i/-2i

=-1

finally a+ib form of given expression is -1+0i

here a=-1,b= 0 & a is real part,b is imaginary part.

Answer 2

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$given \: complex \: number \: is \\ z = ( \frac{1 - i {}^{3} }{1 + i {}^{3} } \: ) {}^{2} \\ \\ = ( \frac{1 - ( - i)}{1 + ( - i)} \: ) {}^{2} \\ \\ = ( \frac{1 + i}{1 - i} \: ) {}^{2} \\ \\ = ( \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} \: ) {}^{2} \\ \\ = ( \frac{1 + i {}^{2} + 2i}{1 - i {}^{2} } \: ) {}^{2} \\ \\ = ( \frac{1 - 1 + 2i}{1 - ( - 1)} \: ) {}^{2} \\ \\ = ( \frac{ 2i}{2} ) {}^{2} \\ \\ = i {}^{2} \\ \\ = - 1 \\ \\ a + ib = - 1 + 0i$