જો $i = \sqrt { - 1} $, તો $1 + {i^2} + {i^3} - {i^6} + {i^8}$ = . . .
$2 - i$
$1$
$3$
$ - 1$
(a) $1 + {i^2} + {i^3} – {i^6} + {i^8} = 1 – 1 – i + 1 + 1 = 2 – i$.
જો $x,y \in R$ અને $(x + iy)(3 + 2i) = 1 + i$, તો $(x,\,y)$ મેળવો.
${i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}}\,,\,(n \in N)$ .= . . .
જો $\,\left| \begin{array}{l}\,6i\,\,\,\,\, – 3i\,\,\,\,\,\,\,\,\,1\\\,\,4\,\,\,\,\,\,\,\,\,3i\,\,\,\,\,\, – 1\\\,20\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,i\end{array} \right|\,$=$x + iy$, તો $(x, y) = . . .$
જો $z = \frac{{7 – i}}{{3 – 4i}}$ તો ${z^{14}} = $
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