Bissoy.com
Doctors Medicines MCQs Q&A

What is i1000 + i1001 + i1002 + i1003 equal to (where i \(= \sqrt { - 1}\))?

What is i1000 + i1001 + i1002 + i1003 equal to (where i \(= \sqrt { - 1}\))? Correct Answer 0

Concept:

Formulas used:

  • i = √(-1)
  • i2 = -1
  • i3 = -i × i2 = -i
  • i4 = (i2)2 = (-1)2 = 1

 

Calculation:

To find: i1000 + i1001 + i1002 + i1003

We know that i4 = 1. So, 1000/4 = 250.

= (i4)250 + i × (i4)250 + i2 × (i4)250 + i3 × (i4)250

= 1 + i + i2 + i3

= 1 + i + (-1) + (-i)

= 0

Related Questions

Which of the following is TRUE? \({\rm{I}}.{\rm{\;}}\sqrt[3]{{11}} > \sqrt 7 > \sqrt[4]{{45}}\) \({\rm{II}}.{\rm{\;}}\sqrt 7 > \sqrt[3]{{11}} > \sqrt[4]{{45}}\) \({\rm{III}}.{\rm{\;}}\sqrt 7 > \sqrt[4]{{45}} > \sqrt[3]{{11}}\) \({\rm{IV}}.{\rm{\;}}\sqrt[4]{{45}} > \sqrt 7 > \sqrt[3]{{11}}\)
Which of the following is TRUE? \({\rm{I}}.{\rm{\;}}\frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt 5 }}\) \({\rm{II}}.\;\frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt 5 }}\) \({\rm{III}}.\;\frac{1}{{\sqrt 5 }} > \frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt[4]{{29}}}}\) \({\rm{IV}}.{\rm{\;}}\frac{1}{{\sqrt 5 }} > \frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt[3]{{12}}}}\)
Simplify : $$\frac{1}{{\sqrt 3 + \sqrt 4 }} \,+ $$   $$\frac{1}{{\sqrt 4 + \sqrt 5 }} \,+ $$   $$\frac{1}{{\sqrt 5 + \sqrt 6 }} \,+ $$   $$\frac{1}{{\sqrt 6 + \sqrt 7 }} \,+ $$   $$\frac{1}{{\sqrt 7 + \sqrt 8 }}\, + $$   $$\frac{1}{{\sqrt 8 + \sqrt 9 }} = ?$$
$$\frac{1}{{\left( {\sqrt 9 - \sqrt 8 } \right)}} \, - $$   $$\frac{1}{{\left( {\sqrt 8 - \sqrt 7 } \right)}} \, + $$   $$\frac{1}{{\left( {\sqrt 7 - \sqrt 6 } \right)}} \, - $$   $$\frac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}} \, + $$   $$\frac{1}{{\left( {\sqrt 5 - \sqrt 4 } \right)}}$$   is equal to ?
Which of the following statement(s) is/are TRUE? I. \(\sqrt{1}+ \sqrt{2} + \sqrt{3} + \sqrt{4}+ \sqrt{5} + \sqrt{6} > 10\) II. \(\sqrt{(10)} + \sqrt{(12)} + \sqrt{(14)} > 3 \sqrt{(12)}\)
If the value of \(\frac{{3x\sqrt y + 2y\sqrt x }}{{3x\sqrt y - 2y\sqrt x }} - \frac{{3x\sqrt y - 2y\sqrt x }}{{3x\sqrt y + 2y\sqrt x }}\) is same as that of \(\sqrt x\sqrt y\), then which of the following relations between x and y is correct?
Which of the following statement(s) is/are TRUE? (I) \(\sqrt {144} \times \sqrt {36} < \sqrt[3]{{125}} \times \sqrt {121}\) (II) \(\sqrt {324} + \sqrt {49} > \sqrt[3]{{216}} \times \sqrt 9\)
Determined the value of $$\frac{1}{{\sqrt 1 + \sqrt 2 }}{\text{ + }}$$  $$\frac{1}{{\sqrt 2 + \sqrt 3 }}\, + $$   $$\frac{1}{{\sqrt 3 + \sqrt 4 }}\, + $$   $$...... + $$   $$\frac{1}{{\sqrt {120} + \sqrt {121} }}{\text{ = ?}}$$
If $$a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }},$$   $$b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$$   then the value of $${a^2} + {b^2}$$   would be = ?
The value of $$\frac{1}{{\sqrt {3.25} + \sqrt {2.25} }}$$    $$ +\, \frac{1}{{\sqrt {4.25} + \sqrt {3.25} }}$$    $$ +\, \frac{1}{{\sqrt {5.25} + \sqrt {4.25} }}$$    $$ +\, \frac{1}{{\sqrt {6.25} + \sqrt {5.25} }}$$    is = ?