Advertisements
Advertisements
प्रश्न
Find the modulus of \[\frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}\].
उत्तर
\[\frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}\]
\[ = \frac{\left( 1 + i \right)\left( 1 + i \right) - \left( 1 - i \right)\left( 1 - i \right)}{\left( 1 - i \right)\left( 1 + i \right)}\]
\[ = \frac{1 + i^2 + 2i - 1 - i^2 + 2i}{1^2 - i^2}\]
\[ = \frac{4i}{2} \left ( \because i^2 = - 1 \right)\]
\[ = 2i\]
\[ \therefore \left| 2i \right| = \sqrt{0^2 + 2^2}\]
\[ = 2 \left( \because \left| a + bi \right| = \sqrt{a^2 + b^2} \right)\]
\[ \Rightarrow \left| \frac{1 + i}{1 - i} - \frac{1 - i}{1 + i} \right| = 2\]
APPEARS IN
संबंधित प्रश्न
Find the modulus and argument of the complex number `(1 + 2i)/(1-3i)`
Find the modulus of `(1+i)/(1-i) - (1-i)/(1+i)`
Find the conjugate of the following complex number:
4 − 5 i
Find the conjugate of the following complex number:
\[\frac{1}{3 + 5i}\]
Find the conjugate of the following complex number:
\[\frac{(1 + i)(2 + i)}{3 + i}\]
Find the conjugate of the following complex number:
\[\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)}\]
Find the modulus and argument of the following complex number and hence express in the polar form:
1 + i
Find the modulus and argument of the following complex number and hence express in the polar form:
1 − i
Find the modulus and argument of the following complex number and hence express in the polar form:
\[\frac{1 - i}{1 + i}\]
Find the modulus and argument of the following complex number and hence express in the polar form:
\[\frac{1}{1 + i}\]
Find the modulus and argument of the following complex number and hence express in the polar form:
\[\frac{1 + 2i}{1 - 3i}\]
If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, prove that \[\arg\left( \frac{z_1}{z_4} \right) + \arg\left( \frac{z_2}{z_3} \right) = 0\].
Write the conjugate of \[\frac{2 - i}{\left( 1 - 2i \right)^2}\] .
If (1 + i) (1 + 2i) (1 + 3i) .... (1 + ni) = a + ib, then 2.5.10.17.......(1+n2)=
If \[\frac{( a^2 + 1 )^2}{2a - i} = x + iy, \text { then } x^2 + y^2\] is equal to
If \[x + iy = (1 + i)(1 + 2i)(1 + 3i)\],then x2 + y2 =
If \[\frac{1 - ix}{1 + ix} = a + ib\] then \[a^2 + b^2\]
If |z2 – 1| = |z|2 + 1, then show that z lies on imaginary axis.
The conjugate of the complex number `(1 - i)/(1 + i)` is ______.
If a complex number lies in the third quadrant, then its conjugate lies in the ______.
If z1 = `sqrt(3) + i sqrt(3)` and z2 = `sqrt(3) + i`, then find the quadrant in which `(z_1/z_2)` lies.
What is the conjugate of `(sqrt(5 + 12i) + sqrt(5 - 12i))/(sqrt(5 + 12i) - sqrt(5 - 12i))`?
If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then find arg`(z_1/z_4)` + arg`(z_2/z_3)`.
Solve the system of equations Re(z2) = 0, z = 2.
What is the conjugate of `(2 - i)/(1 - 2i)^2`?
If `(a^2 + 1)^2/(2a - i)` = x + iy, what is the value of x2 + y2?
sinx + icos2x and cosx – isin2x are conjugate to each other for ______.
