Question

$\text{List I}$ has four entries and $\text{List II}$ has five entries. Each entry of $\text{List I}$ is to be correctly matched with one or more than one entries of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Possible value(s) of}\sqrt{i}+\sqrt{-i}\text{is (are)}& \text{(P)}& \sqrt{2}\\ \text{(B)}& \text{If}{z}^3=\overline{z}(z\ne 0),& \text{(Q)}& i\\ & \text{then possible values of}z\text{is/are}& & \\ \text{(C)}& 1+\frac{1}{4}+\frac{1\cdot 3}{4\cdot 8}+\frac{1\cdot 3\cdot 5}{4\cdot 8\cdot 12}+\cdots \cdots \infty & \text{(R)}& \sqrt{2}i\\ \text{(D)}& \frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3}+\cdots \cdots \infty & \text{(S)}& \frac{1}{2}\\ & & \text{(T)}& \frac{13}{36}\end{array}$

Which of the following is the only CORRECT combination?

• A. $\left(A\right)\to \left(P\right),\left(R\right);\left(B\right)\to \left(Q\right)$
• B. $\left(A\right)\to \left(P\right),\left(Q\right);\left(B\right)\to \left(R\right)$
• C. $\left(B\right)\to \left(P\right),\left(Q\right);\left(B\right)\to \left(S\right)$
• D. $\left(A\right)\to \left(Q\right),\left(R\right);\left(B\right)\to \left(P\right)$

Answer 1

The correct option is A $\left(A\right)\to \left(P\right),\left(R\right);\left(B\right)\to \left(Q\right)$

$\left(A\right)$
$\sqrt{i}=\sqrt{0+1\cdot i}\Rightarrow \sqrt{i}=\pm \{\sqrt{\frac{1}{2}(\sqrt{0^2+1^2}+0)}+i\sqrt{\frac{1}{2}(\sqrt{0^2+1^2}-0)}\}\Rightarrow \sqrt{i}=\pm \frac{1}{\sqrt{2}}(1+i)$
Similarly,
$\sqrt{-i}=\pm \frac{1}{\sqrt{2}}(1-i)$

$\therefore \text{Possible values of}\sqrt{i}+\sqrt{-i}\text{are}\pm \sqrt{2},\pm \sqrt{2}i$
$\left(A\right)\to \left(P\right),\left(R\right)$

$\left(B\right)$
$z^3=\overline{z}\Rightarrow |z^3|=|\overline{z}|\Rightarrow |z{|}^3=|z|\Rightarrow |z|(|z{|}^2-1)=0\Rightarrow |z{|}^2=1(\because z\ne 0)\Rightarrow |z|=1(\because |z|>0)$
Now,
$z^3=\overline{z}\Rightarrow z^4=z\overline{z}=|z{|}^2=1\Rightarrow z^4=\cos 2n\pi +i\sin 2n\pi \Rightarrow z=\cos \frac{2n\pi }{4}+i\sin \frac{2n\pi }{4}\Rightarrow z=\cos \frac{n\pi }{2}+i\sin \frac{n\pi }{2}$
Putting $n=0,1,2,3$
$\Rightarrow z=\pm 1,\pm i$
$\left(B\right)\to \left(Q\right)$