Class 11Mathematics4-1.Complex numbers Integral power of iota, Algebraic operations and Equality of complex numbers
EasyIf $i = \sqrt{-1}$,then $1 + i^2 + i^3 - i^6 + i^8$ is equal to
- A$2 - i$
- B$1$
- C$3$
- D$-1$
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Similar Questions
Express the following expression in the form of $a+ib$:
$\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$
$\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$
Medium
View SolutionIf $\left(\frac{1-i}{1+i}\right)^{96}=a+ib$,then $(a, b)$ is
MediumKCET 2018
View SolutionThe value of the series $1 + i^2 + i^4 + i^6 + ..... + i^{2n}$ is:
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View Solution$\left( \frac{1}{1 - 2i} + \frac{3}{1 + i} \right) \left( \frac{3 + 4i}{2 - 4i} \right) = $
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View SolutionThe real part of the complex number $z = \frac{5+2i}{2-5i} - \frac{3-4i}{4+3i} - \frac{1}{i}$ is
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