{(1 + i)^8} + {(1 - i)^8} = . . .

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4-1.Complex numbers

easy

${(1 + i)^8} + {(1 - i)^8}$ = . . .

$16$

$-16$

$32$

$-32$

Solution

(c) $\,{(1 + i)^2} = \,1 + {i^2} + 2i = 2i$and ${(1 – i)^2} = 1 + {i^2} – 2i = – 2i$
$ \Rightarrow {(1 + i)^8} + {(1 – i)^8} = {(2i)^4} + {( – 2i)^4}$$ = {2^4}({i^4} + {i^4})$
$ = {2^5} = 32.$

Standard 11

Mathematics

જો $\frac{{3x + 2iy}}{{5i – 2}} = \frac{{15}}{{8x + 3iy}}$ તો

medium

જો ${\left( {\frac{{1 – i}}{{1 + i}}} \right)^{100}} = a + ib$ તો . . .

easy

જો $(a+i b)(c+i d)(e+i f)(g+i h)=A+i B,$ હોય તો, બતાવો કે $\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\left(e^{2}+f^{2}\right)\left(g^{2}+h^{2}\right)=A^{2}+B^{2}$

medium

જો ${z_1} = 1 – i$ અને ${z_2} = – 2 + 4i$, તો ${\mathop{\rm Im}\nolimits} \left( {\frac{{{z_1}{z_2}}}{{{z_1}}}} \right) = $

easy

${i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}}\,,\,(n \in N)$ .= . . .

normal

${(1 + i)^8} + {(1 - i)^8}$ = . . .

Solution

Similar Questions

જો $\frac{{3x + 2iy}}{{5i – 2}} = \frac{{15}}{{8x + 3iy}}$ તો

જો ${\left( {\frac{{1 – i}}{{1 + i}}} \right)^{100}} = a + ib$ તો . . .

જો $(a+i b)(c+i d)(e+i f)(g+i h)=A+i B,$ હોય તો, બતાવો કે $\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\left(e^{2}+f^{2}\right)\left(g^{2}+h^{2}\right)=A^{2}+B^{2}$

જો ${z_1} = 1 – i$ અને ${z_2} = – 2 + 4i$, તો ${\mathop{\rm Im}\nolimits} \left( {\frac{{{z_1}{z_2}}}{{{z_1}}}} \right) = $

${i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}}\,,\,(n \in N)$ .= . . .

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