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4-1.Complex numbers
medium
If $z = \frac{{7 - i}}{{3 - 4i}}$ then ${z^{14}} = $
A
${2^7}$
B
${2^7}i$
C
${2^{14}}i$
D
$ - {2^7}i$
Solution
(d) $z = \frac{{7 – i}}{{3 – 4i}} \times \frac{{3 + 4i}}{{3 + 4i}}$=$\frac{{21 + 25i + 4}}{{16 + 9}} = \frac{{25\,(1 + i)}}{{25}}$ = $(1 + i)$
${z^{14}} = {(1 + i)^{14}} = {[{(1 + i)^2}]^7}$= ${(2i)^7} = {2^7}{i^7} = – {2^7}i$.
Standard 11
Mathematics