मराठी

If Z = a + I B Lies in Third Quadrant, Then ¯ Z Z Also Lies in Third Quadrant If

Advertisements
Advertisements

प्रश्न

If \[z = a + ib\]  lies in third quadrant, then \[\frac{\bar{z}}{z}\] also lies in third quadrant if

पर्याय

  • \[a > b > 0\]

  • \[a < b < 0\]

  • \[b < a < 0\]

  • \[b > a > 0\]

MCQ
Advertisements

उत्तर

Since, \[z = a + ib\] lies in third quadrant. \[\Rightarrow a < 0 \text { and } b < 0 . . . . (1)\]

Now,

\[\frac{\bar{z}}{z} = \frac{\bar{{a + ib}}}{a + ib}\]

\[ = \frac{a - ib}{a + ib}\]

\[ = \frac{a - ib}{a + ib} \times \frac{a - ib}{a - ib}\]

\[ = \frac{a^2 + i^2 b^2 - 2abi}{a^2 - i^2 b^2}\]

\[ = \frac{a^2 - b^2 - 2abi}{a^2 + b^2}\]

Since, 

\[\frac{\bar{z}}{z}\] also lies in third quadrant.

\[\Rightarrow a^2 - b^2 < 0\]

\[ \Rightarrow (a - b)(a + b) < 0\]

\[ \Rightarrow a - b > 0 \text { and  }a + b < 0\]

\[ \Rightarrow a > b . . . . (2)\]

From (1) and (2),

\[b < a < 0\]

Hence, the correct option is (c).

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Complex Numbers - Exercise 13.6 [पृष्ठ ६६]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 13 Complex Numbers
Exercise 13.6 | Q 37 | पृष्ठ ६६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Express the given complex number in the form a + ib: `(1/5 + i 2/5) - (4 + i 5/2)`


If a + ib  = `(x + i)^2/(2x^2 + 1)` prove that a2 + b= `(x^2 + 1)^2/(2x + 1)^2`


Evaluate the following:

(ii) i528


Find the value of the following expression:

i + i2 + i3 + i4


Find the value of the following expression:

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


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

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


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

\[(1 + i)(x + iy) = 2 - 5i\]


Find the multiplicative inverse of the following complex number:

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


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.


Solve the equation \[\left| z \right| = z + 1 + 2i\].


Write (i25)3 in polar form.


Express \[\sin\frac{\pi}{5} + i\left( 1 - \cos\frac{\pi}{5} \right)\] in polar form.


If n is any positive integer, write the value of \[\frac{i^{4n + 1} - i^{4n - 1}}{2}\].


Find the principal argument of \[\left( 1 + i\sqrt{3} \right)^2\] .


If \[\left| z - 5i \right| = \left| z + 5i \right|\] , then find the locus of z.


If \[\frac{\left( a^2 + 1 \right)^2}{2a - i} = x + iy\] find the value of  \[x^2 + y^2\].


Write the sum of the series \[i + i^2 + i^3 + . . . .\] upto 1000 terms.


The value of \[(1 + i)(1 + i^2 )(1 + i^3 )(1 + i^4 )\] is.


The polar form of (i25)3 is


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


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


If \[z = \frac{1}{1 - cos\theta - i sin\theta}\] then Re (z) =


If \[x + iy = \frac{3 + 5i}{7 - 6i},\]  then y =


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


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


The value of \[(1 + i )^4 + (1 - i )^4\] is


The complex number z which satisfies the condition \[\left| \frac{i + z}{i - z} \right| = 1\] lies on


If z is a complex numberthen


Simplify : `sqrt(-16) + 3sqrt(-25) + sqrt(-36) - sqrt(-625)`


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:

`("i"(4 + 3"i"))/((1 - "i"))`


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


Evaluate the following : i93  


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


Evaluate the following : i–888 


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).


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


Show that `(-1+sqrt3i)^3` is a real number.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×