मराठी

If I2 = −1, Then the Sum I + I2 + I3 +... Upto 1000 Terms is Equal to - Mathematics

Advertisements
Advertisements

प्रश्न

If i2 = −1, then the sum i + i2 + i3 +... upto 1000 terms is equal to

पर्याय

  • 1

  • −1

  • i

  • 0

MCQ
Advertisements

उत्तर

0

\[i + i^2 + i^3 + i^4 . . . i^{1000} \]

\[ i + i^2 + i^3 + i^4 [ \because i^2 = - 1, i^3 = - i \text { and } i^4 = 1]\]

\[ = i - 1 - i + 1 \]

\[ = 0 \]

\[\text { Similarly, the sum of the next four terms of the series will be equal to 0 . This is because the powers of i follow a cyclicity of 4 } . \]

\[\text { Hence, the sum of all terms, till 1000, will be zero } . \]

\[i + i^2 + i^3 + i^4 . . . i^{1000} = 0\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Complex Numbers - Exercise 13.6 [पृष्ठ ६४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 13 Complex Numbers
Exercise 13.6 | Q 7 | पृष्ठ ६४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Express the given complex number in the form a + ib: (1 – i) – (–1 + i6)


Express the given complex number in the form a + ib: `(1/3 + 3i)^3`


Evaluate: `[i^18 + (1/i)^25]^3`


Find the value of the following expression:

i5 + i10 + i15


Find the value of the following expression:

\[\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}}\]


Find the value of the following expression:

1+ i2 + i4 + i6 + i8 + ... + i20


Find the value of the following expression:

(1 + i)6 + (1 − i)3


Express the following complex number in the standard form a + i b:

\[\frac{3 + 2i}{- 2 + i}\]


Express the following complex number in the standard form a + i b:

\[\frac{(1 + i)(1 + \sqrt{3}i)}{1 - i}\] .


Express the following complex number in the standard form a + i b:

\[\frac{(1 - i )^3}{1 - i^3}\]


Express the following complex number in the standard form a + i b:

\[\left( \frac{1}{1 - 4i} - \frac{2}{1 + i} \right)\left( \frac{3 - 4i}{5 + i} \right)\]


Find the real value of x and y, if

\[(x + iy)(2 - 3i) = 4 + i\]


Find the real value of x and y, if

\[(3x - 2iy)(2 + i )^2 = 10(1 + i)\]


If \[x + iy = \frac{a + ib}{a - ib}\] prove that x2 + y2 = 1.


Evaluate the following:

\[x^4 - 4 x^3 + 4 x^2 + 8x + 44,\text {  when } x = 3 + 2i\]


For a positive integer n, find the value of \[(1 - i )^n \left( 1 - \frac{1}{i} \right)^n\].


What is the smallest positive integer n for which \[\left( 1 + i \right)^{2n} = \left( 1 - i \right)^{2n}\] ?


If z1 and z2 are two complex numbers such that \[\left| z_1 \right| = \left| z_2 \right|\] and arg(z1) + arg(z2) = \[\pi\] then show that \[z_1 = - \bar{{z_2}}\].


Write 1 − i in polar form.


Write the least positive integral value of n for which  \[\left( \frac{1 + i}{1 - i} \right)^n\] is real.


Find the real value of a for which \[3 i^3 - 2a i^2 + (1 - a)i + 5\] is real.


If \[z = \frac{- 2}{1 + i\sqrt{3}}\],then the value of arg (z) is


The least positive integer n such that \[\left( \frac{2i}{1 + i} \right)^n\] is a positive integer, is.

 

The argument of \[\frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}\] is


If \[z = \frac{1 + 2i}{1 - (1 - i )^2}\], then arg (z) equal


The value of (i5 + i6 + i7 + i8 + i9) / (1 + i) is


Find a and b if (a+b) (2 + i) = b + 1 + (10 + 2a)i


Find a and b if abi = 3a − b + 12i


Find a and b if (a + ib) (1 + i) = 2 + i


Express the following in the form of a + ib, a, b ∈ R, i = `sqrt(−1)`. State the values of a and b:

(1 + i)(1 − i)−1 


Express the following in the form of a + ib, a, b ∈ R i = `sqrt(−1)`. State the values of a and b:

`(2 + sqrt(-3))/(4 + sqrt(-3))`


Show that `(-1 + sqrt(3)"i")^3` is a real number


Evaluate the following : i35 


Evaluate the following : i888 


Show that 1 + i10 + i20 + i30 is a real number


Answer the following:

Show that z = `5/((1 - "i")(2 - "i")(3 - "i"))` is purely imaginary number.


If `((1 - i)/(1 + i))^100` = a + ib, then find (a, b).


Find the value of `(i^(592) + i^(590) + i^(588) + i^(586) + i^(584))/(i^(582) + i^(580) + i^(578) + i^(576) + i^(574))`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×