मराठी

If Z = 1 + 7 I ( 2 − I ) 2 , Then - Mathematics

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • \[\left| z \right| = 2\]

  • \[\left| z \right| = \frac{1}{2}\]

  • amp (z) = \[\frac{\pi}{4}\]

  •  amp (z) = \[\frac{3\pi}{4}\]

MCQ

उत्तर

amp (z) = \[\frac{3\pi}{4}\]

\[z = \frac{1 + 7i}{\left( 2 - i \right)^2}\]

\[ \Rightarrow z = \frac{1 + 7i}{4 + i^2 - 4i}\]

\[ \Rightarrow z = \frac{1 + 7i}{4 - 1 - 4i} \left[ \because i^2 = - 1 \right]\]

\[ \Rightarrow z = \frac{1 + 7i}{3 - 4i}\]

\[ \Rightarrow z = \frac{1 + 7i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i}\]

\[ \Rightarrow z = \frac{3 + 4i + 21i + 28 i^2}{9 - 16 i^2}\]

\[ \Rightarrow z = \frac{3 - 28 + 25i}{9 + 16}\]

\[ \Rightarrow z = \frac{- 25 + 25i}{25}\]

\[ \Rightarrow z = - 1 + i\]

\[\tan \alpha = \left| \frac{Im\left( z \right)}{Re\left( z \right)} \right|\]

\[ = 1\]

\[ \Rightarrow \alpha = \frac{\pi}{4}\]

\[\text { Since, z lies in the second quadrant }. \]

\[\text { Therefore, amp } (z) = \pi - \alpha\]

\[ = \pi - \frac{\pi}{4} \]

\[ = \frac{3\pi}{4} \]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Complex Numbers - Exercise 13.6 [पृष्ठ ६५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 13 Complex Numbers
Exercise 13.6 | Q 29 | पृष्ठ ६५

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:

`[(1/3 + i 7/3) + (4 + i 1/3)] -(-4/3 + i)`


Express the given complex number in the form a + ib: `(1/3 + 3i)^3`


Evaluate the following:

i457


Find the value of the following expression:

i49 + i68 + i89 + i110


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:

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


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

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


If \[z_1 = 2 - i, z_2 = 1 + i,\text {  find } \left| \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} \right|\]


For a positive integer n, find the value of \[(1 - i )^n \left( 1 - \frac{1}{i} \right)^n\].


Write (i25)3 in polar form.


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


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


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


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


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


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


The polar form of (i25)3 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 \[\frac{x}{a} + \frac{y}{b} =\]


\[\text { If  }z = 1 - \text { cos }\theta + i \text { sin }\theta, \text { then } \left| z \right| =\]


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


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


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


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


Find a and b if (a – b) + (a + b)i = a + 5i


Find a and b if `1/("a" + "ib")` = 3 – 2i


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


If z1 = 3 – 2i and z2 = –1 + 3i, then Im(z1z2) = ______.


State true or false for the following:

If a complex number coincides with its conjugate, then the number must lie on imaginary axis.


If a = cosθ + isinθ, find the value of `(1 + "a")/(1 - "a")`.


If w is a complex cube-root of unity, then prove the following

(w2 + w − 1)3 = −8


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×