English

1 + 2 I + 3 I 2 1 − 2 I + 3 I 2 Equals

Advertisements
Advertisements

Question

\[\frac{1 + 2i + 3 i^2}{1 - 2i + 3 i^2}\] equals

Options

  • i

  • -1

  • \[-\]i

  • 4

MCQ
Advertisements

Solution

\[-\]i

\[\text { Let z } = \frac{1 + 2i + 3 i^2}{1 - 2i + 3 i^2}\]

\[ \Rightarrow z = \frac{1 + 2i - 3}{1 - 2i - 3}\]

\[ \Rightarrow z=\frac{- 2 + 2i}{- 2 - 2i}\times$\frac{- 2 + 2i}{- 2 + 2i}\]

\[ \Rightarrow z=\frac{\left( - 2 + 2i \right)^2}{\left( - 2 \right)^2 - \left( 2i \right)^2}\]

\[ \Rightarrow z=\frac{4 + 4 i^2 - 8i}{4 + 4}\]

\[ \Rightarrow z =\frac{4 - 4 - 8i}{8}\]

\[ \Rightarrow z=\frac{- 8i}{8}\]

\[ \Rightarrow z =-i\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Complex Numbers - Exercise 13.6 [Page 66]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 13 Complex Numbers
Exercise 13.6 | Q 34 | Page 66

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Evaluate the following:

 \[i^{30} + i^{40} + i^{60}\]


Find the value of the following expression:

i5 + i10 + i15


Find the value of the following expression:

1+ i2 + i4 + i6 + i8 + ... + i20


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:

\[\frac{1 - i}{1 + i}\]


Find the real value of x and y, if

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


Find the real value of x and y, if `((1+i)x-2i)/(3+i) + ((2-3i)y+i)/(3-i) = i, xy ∈ R, i = sqrt-1`


Find the multiplicative inverse of the following complex number:

\[(1 + i\sqrt{3} )^2\]


Find the real values of θ for which the complex number \[\frac{1 + i cos\theta}{1 - 2i cos\theta}\]  is purely real.


If z1 is a complex number other than −1 such that \[\left| z_1 \right| = 1\] and \[z_2 = \frac{z_1 - 1}{z_1 + 1}\] then show that the real parts of z2 is zero.


If \[\frac{z - 1}{z + 1}\] is purely imaginary number (\[z \neq - 1\]), find the value of \[\left| z \right|\].


If π < θ < 2π and z = 1 + cos θ + i sin θ, then write the value of \[\left| z \right|\] .


Write the argument of −i.


Find z, if \[\left| z \right| = 4 \text { and }\arg(z) = \frac{5\pi}{6} .\]


Write the argument of \[\left( 1 + i\sqrt{3} \right)\left( 1 + i \right)\left( \cos\theta + i\sin\theta \right)\].

Disclaimer: There is a misprinting in the question. It should be  \[\left( 1 + i\sqrt{3} \right)\]  instead of \[\left( 1 + \sqrt{3} \right)\].


If\[z = \cos\frac{\pi}{4} + i \sin\frac{\pi}{6}\], then


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

 

If z is a non-zero complex number, then \[\left| \frac{\left| z \right|^2}{zz} \right|\] is equal to


If (x + iy)1/3 = a + ib, then \[\frac{x}{a} + \frac{y}{b} =\]


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


If \[z = \frac{1 + 7i}{(2 - i )^2}\] , then


The amplitude of \[\frac{1}{i}\] is equal to


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


If z is a complex numberthen


Which of the following is correct for any two complex numbers z1 and z2?

 


If the complex number \[z = x + iy\] satisfies the condition \[\left| z + 1 \right| = 1\], then z lies on


Find a and b if a + 2b + 2ai = 4 + 6i


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

(1 + 2i)(– 2 + i)


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

(2 + 3i)(2 – 3i)


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

`(4"i"^8 - 3"i"^9 + 3)/(3"i"^11 - 4"i"^10 - 2)`


Evaluate the following : i35 


Evaluate the following : i116 


Evaluate the following : `1/"i"^58`


If `((1 + "i"sqrt3)/(1 - "i"sqrt3))^"n"` is an integer, then n is ______.


The real value of θ for which the expression `(1 + i cos theta)/(1 - 2i cos theta)` is a real number is ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×