मराठी

If z1 and z2 both satisfy z+z¯=2|z-1| arg(z1-z2)=π4, then find ImIm(z1+z2). - Mathematics

Advertisements
Advertisements

प्रश्न

If z1 and z2 both satisfy `z + barz = 2|z - 1|` arg`(z_1 - z_2) = pi/4`, then find `"Im" (z_1 + z_2)`.

बेरीज
Advertisements

उत्तर

Let z = x + iy, z1 = x1 + iy1 and z2 = x2 + iy2 .

Then `z + barz = 2|z - 1|`

⇒ (x + iy) + (x – iy) = `2|x - 1 + "i"y|`

⇒ 2x = 1 + y2    .......(1)

Since z1 and z2 both satisfy (1), we have

`2x_1 = 1 + y_1^2 .....` and `2x_2 = 1 + y_2^2`

⇒ `2(x_1 - x_2) = (y_1 + y_2)(y_1 - y_2)`

⇒ 2 = `(y_1 + y_2) ((y_1 - y_2)/(x_1 - x_2))`  ......(2)

Again `z_1 - "z"_2 = (x_1 - x_2) + "i"(y_"i" - y_2)`

Therefore, tanθ = `(y_1 - y_2)/(x_1 - x_2)`, where θ = arg`("z"_1 - "z"_2)`

⇒ `tan  pi/4 = (y_1 - y_2)/(x_1 - x_2)`  ......`("Since"  theta = pi/4)`

i.e., 1 = `(y_1 - y_2)/(x_1 - x_2)`

From (2), We get 2 = y1 + y2 i.e., `"Im" ("z"_1 + "z"_2)` = 2

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Complex Numbers and Quadratic Equations - Solved Examples [पृष्ठ ८३]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 5 Complex Numbers and Quadratic Equations
Solved Examples | Q 15 | पृष्ठ ८३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Express the given complex number in the form a + ib: i9 + i19


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`


Evaluate the following:

i457


Evaluate the following:

(ii) i528


Evaluate the following:

\[( i^{77} + i^{70} + i^{87} + i^{414} )^3\]


Find the value of the following expression:

i + i2 + i3 + i4


Find the value of the following expression:

i5 + i10 + i15


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

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


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`


Evaluate the following:

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


Solve the system of equations \[\text { Re }\left( z^2 \right) = 0, \left| z \right| = 2\].


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.


If z1z2z3 are complex numbers such that \[\left| z_1 \right| = \left| z_2 \right| = \left| z_3 \right| = \left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right| = 1\] then find the value of \[\left| z_1 + z_2 + z_3 \right|\] .


Find the number of solutions of \[z^2 + \left| z \right|^2 = 0\].


Write (i25)3 in polar form.


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.


If \[\left| z + 4 \right| \leq 3\], then find the greatest and least values of \[\left| z + 1 \right|\].


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


The polar form of (i25)3 is


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


If \[z = \left( \frac{1 + i}{1 - i} \right)\] then z4 equals


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


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


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


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


A real value of x satisfies the equation  \[\frac{3 - 4ix}{3 + 4ix} = a - ib (a, b \in \mathbb{R}), if a^2 + b^2 =\]


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

 


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:

`(3 + 2"i")/(2 - 5"i") + (3 -2"i")/(2 + 5"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)−3 


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


Answer the following:

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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×