Question

Statement I Multiplicative inverse of $2-3 i$ is $2+3 i$.
Statement II Simplest form of $\frac{5+\sqrt{2} i}{1-\sqrt{2} i}$ is $1+2 \sqrt{2} i$. Statement III Simplest form of $i^{-35}$ is $-i$.

• A. Statement I is correct
• B. Statement II is correct
• C. Statement III is correct
• D. All are correct

Answer 1

Answer: (B)

I. Let $z=2-3 i$
Then, $\bar{z}=2+3 i$ and $|z|^2=2^2+(-3)^2=13$
Therefore, the multiplicative inverse of $2-3 i$ is
$z^{-1}=\frac{\bar{z}}{|z|^2}=\frac{2+3 i}{13}=\frac{2}{13}+\frac{3}{13} i$
The above working can be reproduced in the following manner also,
$z^{-1}=\frac{1}{2-3 i}=\frac{2+3 i}{(2-3 i)(2+3 i)}$
$=\frac{2+3 i}{2^2-(3 i)^2}=\frac{2+3 i}{13}=\frac{2}{13}+\frac{3}{13} i$
II. We have,
$\frac{5+\sqrt{2} i}{1-\sqrt{2} i} =\frac{5+\sqrt{2} i}{1-\sqrt{2} i} \times \frac{1+\sqrt{2} i}{1+\sqrt{2} i} $
$ =\frac{5+5 \sqrt{2} i+\sqrt{2} i-2}{1-(\sqrt{2} i)^2}$
$ =\frac{3+6 \sqrt{2} i}{1+2}=\frac{3(1+2 \sqrt{2} i)}{3}=1+2 \sqrt{2} i$
III. $i^{-35}=\frac{1}{i^{35}}=\frac{1}{\left(i^2\right)^{17} i}=\frac{1}{-i} \times \frac{i}{i}=\frac{i}{-i^2}=i$