The polar form of i 253 isA- cos-2- i sin-2B- cos- i sin-C- cos- i sin-D- cos-2- i sin-2

The correct option is D cos π/2 i sin π/2 (i^25)^3 =(i)^75 =(i)^4×18+3=(i)^3 = i (∵ i^4=1) Let z=0 i Since, the point (0, 1) lies on the negative ...

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Vector Component

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Question

The polar form of $(i^{25})^{3}$ is

$c o s \frac{π}{2} + i s i n \frac{π}{2}$

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$c o s π + i s i n π$

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$c o s π + i s i n π$

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$c o s \frac{π}{2} - i s i n \frac{π}{2}$

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\cos \frac{π}{2} + i \sin \frac{π}{2}

\cos \frac{π}{2} - i \sin \frac{π}{2}

The polar form of complex number √3 + i

(c o s \frac{π}{8} + i s i n \frac{π}{8}) \times (c o s \frac{π}{12} + i s i n \frac{π}{12}) \times (c o s \frac{π}{24} + i s i n \frac{π}{24}) \times (c o s \frac{π}{4} + i s i n \frac{π}{4})

{(\frac{1 + c o s \frac{π}{2} - i s i n \frac{π}{2}}{1 + c o s \frac{π}{2} + i s i n \frac{π}{2}})}^{2}

The value of Expression $(c o s \frac{π}{2} + i s i n \frac{π}{2})$

$(c o s \frac{π}{2^{2}} + i s i n \frac{π}{2^{2}})$ ...........to $\infty$ is

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