KCET · Maths · Complex Number
The modulus and amplitude of \(\frac{1+2 i}{1-(1-i)^{2}}\) are
- A \(\sqrt{2}\) and \(\frac{\pi}{6}\)
- B 1 and \(\frac{\pi}{4}\)
- C 1 and 0
- D 1 and \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(C) 1 and 0
Step-by-step Solution
Detailed explanation
Let \(z=\frac{1+2 i}{1-(1-i)^{2}}\)
\(\Rightarrow z=\frac{1+2 i}{1-\left(1+i^{2}-2 i\right)}=\frac{1+2 i}{-i^{2}+2 i}\)
\(=\frac{1+2 i}{1+2 i} \quad\left(\because i^{2}=-1\right)\)
\(\Rightarrow \quad z=1+0 i\)
\(\therefore\) Modulus of \(z=|z|=|1+0 \cdot i|\)
\(=\sqrt{1^{2}+0^{2}}=1\)
and amplitude of \(z \equiv \tan ^{-1}\left(\frac{0}{1}\right)\)
\(=\tan ^{-1}(0)=\tan ^{-1}(\tan 0)\)
\(=0\)
\(\Rightarrow z=\frac{1+2 i}{1-\left(1+i^{2}-2 i\right)}=\frac{1+2 i}{-i^{2}+2 i}\)
\(=\frac{1+2 i}{1+2 i} \quad\left(\because i^{2}=-1\right)\)
\(\Rightarrow \quad z=1+0 i\)
\(\therefore\) Modulus of \(z=|z|=|1+0 \cdot i|\)
\(=\sqrt{1^{2}+0^{2}}=1\)
and amplitude of \(z \equiv \tan ^{-1}\left(\frac{0}{1}\right)\)
\(=\tan ^{-1}(0)=\tan ^{-1}(\tan 0)\)
\(=0\)
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