English

If Z1 and Z2 Are Two Complex Numbers Such that | Z 1 | = | Z 2 | and Arg(Z1) + Arg(Z2) = π Then Show that Z 1 = − ¯ Z 2 .

Advertisements
Advertisements

Question

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 \[z_1 = - \bar{{z_2}}\].

Advertisements

Solution

Let θbe the arg(z1) and θbe the arg(z2).
It is given that

\[\left| z_1 \right| = \left| z_2 \right|\] and arg(z1) + arg(z2) = \[\pi\].

Since, z1 is a complex number.

\[z_1 = \left| z_1 \right|\left( \cos \theta_1 + i\sin \theta_1 \right)\]

\[ = \left| z_2 \right|\left[ \cos\left( \pi - \theta_2 \right) + i\sin\left( \pi - \theta_2 \right) \right]\]

\[ = \left| z_2 \right|\left[ - \cos\left( \theta_2 \right) + i\sin\left( \theta_2 \right) \right]\]

\[ = - \left| z_2 \right|\left[ \cos\left( \theta_2 \right) - i\sin\left( \theta_2 \right) \right]\]

\[ = - \bar{{z_2}}\]

Hence,  

\[z_1 = - \bar{{z_2}}\].

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Complex Numbers - Exercise 13.4 [Page 57]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 13 Complex Numbers
Exercise 13.4 | Q 4 | Page 57

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Express the given complex number in the form a + ib: `(5i) (- 3/5 i)`


Let z1 = 2 – i, z2 = –2 + i. Find `"Im"(1/(z_1barz_1))`


Evaluate the following:

i457


Evaluate the following:

\[i^{37} + \frac{1}{i^{67}}\].


Evaluate the following:

\[i^{49} + i^{68} + i^{89} + i^{110}\]


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`


If \[\left( \frac{1 - i}{1 + i} \right)^{100} = a + ib\] find (a, b).


If \[a = \cos\theta + i\sin\theta\], find the value of \[\frac{1 + a}{1 - a}\].


Evaluate the following:

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


Express the following complex in the form r(cos θ + i sin θ):
1 + i tan α


Write −1 + \[\sqrt{3}\] in polar form .


Write the value of \[\arg\left( z \right) + \arg\left( \bar{z} \right)\].


For any two complex numbers z1 and z2 and any two real numbers a, b, find the value of \[\left| a z_1 - b z_2 \right|^2 + \left| a z_2 + b z_1 \right|^2\].


If n ∈ \[\mathbb{N}\] then find the value of \[i^n + i^{n + 1} + i^{n + 2} + i^{n + 3}\] .


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


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


The polar form of (i25)3 is


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


If a = cos θ + i sin θ, then \[\frac{1 + a}{1 - a} =\]


The principal value of the amplitude of (1 + i) is


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


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


If θ is the amplitude of \[\frac{a + ib}{a - ib}\] , than tan θ =


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


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


If \[f\left( z \right) = \frac{7 - z}{1 - z^2}\] , where \[z = 1 + 2i\] then \[\left| f\left( z \right) \right|\] is


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

 


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


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:

(1 + i)−3 


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

`(- sqrt(5) + 2sqrt(-4)) + (1 -sqrt(-9)) + (2 + 3"i")(2 - 3"i")`


Find the value of `(3 + 2/i) (i^6 - i^7) (1 + i^11)`.


Evaluate the following : i93  


State True or False for the following:

2 is not a complex number.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×