Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS1 Mark
Question
Find the modulus and argument of the complex number $\frac{1+i}{1-i}$
✓
Answer
We have $\frac{1+i}{1-i}=\frac{1+i}{1-i} \times \frac{1+i}{1+i}=\frac{1-1+2 i}{1+1}$ = i = 0 + i
Now, let us put 0 = r cos $\theta$, 1 = r sin $\theta$
Squaring and adding, $r^2 = 1$ i.e., $r = 1$ so that
cos $\theta$ = 0, sin $\theta$ = 1
Therefore, $\theta = \frac {\pi}{2}$
Hence, the modulus of $\frac {1+i}{1-i}$ is 1 and the argument is $\frac {\pi}2.$
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