Question

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