The value of ${(1 + i)^5} \times {(1 - i)^5}$ is
$-8$
$8i$
$8$
$32$
(d) ${(1 + i)^5}{(1 – i)^5} = {(1 – {i^2})^5} = {2^5} = 32$.
Express the following expression in the form of $a+i b$.
$\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2 i})-(\sqrt{3}-i \sqrt{2})}$
Let $S_{1}=\left\{z_{1} \in C:\left|z_{1}-3\right|=\frac{1}{2}\right\}$ and $S_{2}=\left\{z_{2} \in C:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\} . \quad$ Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2}$, the least value of $\left|z_{2}-z_{1}\right|$ is.
$\left( {\frac{1}{{1 – 2i}} + \frac{3}{{1 + i}}} \right)\,\,\left( {\frac{{3 + 4i}}{{2 – 4i}}} \right) = $
If $(1 + i)(1 + 2i)(1 + 3i)…..(1 + ni) = a + ib$, then $2.5.10….$$(1 + {n^2})$ is equal to
Solve the equation $3 x^{2}-4 x+\frac{20}{3}=0$
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