Search any question & find its solution
Question:
Answered & Verified by Expert
The modulus and principal argument of the complex number
$\frac{1+2 i}{1-(1-i)^{2}}$ are respectively
Options:
$\frac{1+2 i}{1-(1-i)^{2}}$ are respectively
Solution:
2307 Upvotes
Verified Answer
The correct answer is:
1,0
We know, if $Z=x+$ iy,$|z|=\sqrt{x^{2}+y^{2}}$
Given, $\frac{1+2 i}{1-(1-i)^{2}}=\frac{1+2 i}{1-\left(1-2 i+i^{2}\right)}=\frac{1+2 i}{1+2 i}=1$
Since, it is purelyreal number, modulus $=1$ and principal argument $=0$
Given, $\frac{1+2 i}{1-(1-i)^{2}}=\frac{1+2 i}{1-\left(1-2 i+i^{2}\right)}=\frac{1+2 i}{1+2 i}=1$
Since, it is purelyreal number, modulus $=1$ and principal argument $=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.