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Asked by Sajjan Last Modified
Answer As a seasoned tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your math question.
To find the modulus of (1+i)(1−i)−(1−i)(1+i)(1−i)(1+i)−(1+i)(1−i), we'll first simplify the expression.
Step 1: Simplify the fractions within the brackets. =(1+i)(1+i)(1−i)(1+i)−(1−i)(1−i)(1+i)(1−i)=(1−i)(1+i)(1+i)(1+i)−(1+i)(1−i)(1−i)(1−i)
Step 2: Expand the numerators and denominators. =1+2i+i21−i2−1−2i+i21−i2=1−i21+2i+i2−1−i21−2i+i2
Step 3: Simplify the terms using i2=−1i2=−1. =1+2i−11+1−1−2i−11+1=1+11+2i−1−1+11−2i−1
=2i2−−2i2=22i−2−2i
=i+i=i+i
=2i=2i
Now, to find the modulus (absolute value) of 2i2i, we simply take the square root of the sum of the squares of its real and imaginary parts.
∣2i∣=02+22=4=2∣2i∣=02+22
=4
=2
So, the modulus of the given expression is 22. If you need further clarification or assistance, feel free to ask! And remember, UrbanPro is always here to help you excel in your studies.
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