Question
Find the modulus of1+i/1-i-1-i/1+i
Answers
GPT-4o mini
To find the modulus of the given expression
\[
\frac{1+i}{1-i} - \frac{1-i}{1+i},
\]
we will simplify the expression step-by-step.
1. Start with the first term \(\frac{1+i}{1-i}\):
\[
\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i - 1}{1^2 - (-1^2)} = \frac{2i}{2} = i.
\]
2. Now simplify the second term \(\frac{1-i}{1+i}\):
\[
\frac{1-i}{1+i} \cdot \frac{1-i}{1-i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1 - 2i + 1}{1^2 - 1^2} = \frac{2 - 2i}{2} = 1 - i.
\]
3. Now we substitute these simplified expressions back into the overall expression:
\[
\frac{1+i}{1-i} - \frac{1-i}{1+i} = i - (1 - i) = i - 1 + i = 2i - 1.
\]
4. Now that we have \(2i - 1\), we find the modulus:
\[
|2i - 1| = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5}.
\]
Thus, the modulus of the original expression is
\[
\boxed{\sqrt{5}}.
\]
\[
\frac{1+i}{1-i} - \frac{1-i}{1+i},
\]
we will simplify the expression step-by-step.
1. Start with the first term \(\frac{1+i}{1-i}\):
\[
\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i - 1}{1^2 - (-1^2)} = \frac{2i}{2} = i.
\]
2. Now simplify the second term \(\frac{1-i}{1+i}\):
\[
\frac{1-i}{1+i} \cdot \frac{1-i}{1-i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1 - 2i + 1}{1^2 - 1^2} = \frac{2 - 2i}{2} = 1 - i.
\]
3. Now we substitute these simplified expressions back into the overall expression:
\[
\frac{1+i}{1-i} - \frac{1-i}{1+i} = i - (1 - i) = i - 1 + i = 2i - 1.
\]
4. Now that we have \(2i - 1\), we find the modulus:
\[
|2i - 1| = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5}.
\]
Thus, the modulus of the original expression is
\[
\boxed{\sqrt{5}}.
\]