Given \[{{a}_{i\text{ }j}}~=\text{ }{{e}^{2ix}}~sin\text{ }x\text{ }j\] Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\times 2}}\] So, the elements in a \[2\text{ }\times \text{ }2\] matrix are...
Integrate the function in
As per the given question, Let I = = = = = = = = =
If z and w are two complex numbers such that |zw| = 1 and arg (z) – arg (w) = π/2, then show that z̅w = – i.
Let z = \[\left| z \right|\text{ }(cos\text{ }{{\theta }_{1}}~+\text{ }I\text{ }sin\text{ }{{\theta }_{1}})\text{ }and\text{ }w\text{ }=\text{ }\left| w \right|\text{ }(cos\text{ }{{\theta...
Write the complex number in polar from.
SOLUTION:
Find the complex number satisfying the equation z + √2 |(z + 1)| + i = 0.
\[z\text{ }+\text{ }\surd 2\text{ }\left| \left( z\text{ }+\text{ }1 \right) \right|\text{ }+\text{ }i\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 1 \right)\] Substituting\[z\text{ }=\text{...
Solve the system of equations Re (z2) = 0, |z| = 2.
\[\begin{array}{*{35}{l}} Re\text{ }({{z}^{2}})\text{ }=\text{ }0,\text{ }\left| z \right|\text{ }=\text{ }2 \\ Let\text{ }z\text{ }=\text{ }x\text{ }+\text{ }iy. \\ Then,\text{ }\left| z...
If for complex numbers z1 and z2, arg (z1) – arg (z2) = 0, then show that |z1 – z2| = |z1| – |z2|.
Let \[{{z}_{1}}~=\text{ }\left| {{z}_{1}} \right|\text{ }\left( cos\text{ }{{\theta }_{1}}~+\text{ }I\text{ }sin\text{ }{{\theta }_{1}} \right)\text{ }and\text{ }{{z}_{2}}~=\text{ }\left| {{z}_{2}}...
If |z1| = |z2| = ….. = |zn| = 1, then show that |z1 + z2 + z3 + …. + zn| = | 1/z1 + 1/z2 + 1/z3 + … + 1/zn|
\[\begin{array}{*{35}{l}} \left| {{z}_{1}} \right|\text{ }=\text{ }\left| {{z}_{2}} \right|\text{ }=\text{ }\ldots \text{ }=\text{ }\left| {{z}_{n}} \right|\text{ }=\text{ }1 \\ \Rightarrow...
If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then find arg(z1/z4) + arg(z2/z3).
z1 and z2 are conjugate complex numbers. The negative side of the real axis \[\begin{array}{*{35}{l}} =\text{ }{{r}_{1}}~\left( cos\text{ }{{\theta }_{1}}~\text{ }-i\text{ }sin\text{ }{{\theta...
If |z1| = 1 (z1 ≠ –1) and z2 = (z1 – 1) / (z + 1), then show that the real part of z2 is zero.
Let z1 = x + iy Therefore, the real part of z2 is zero.
z1 and z2 are two complex numbers such that |z1| = |z2| and arg (z1) + arg (z2) = π, then show that z1 = – z̅2.
According to the question, Let \[{{z}_{1}}~=\text{ }|{{z}_{1}}|\text{ }(cos\text{ }{{\theta }_{1}}~+\text{ }I\text{ }sin\text{ }{{\theta }_{1}})\text{ }and\text{ }{{z}_{2}}~=\text{...
If (z – 1)/(z + 1) is a purely imaginary number (z ≠ –1), then find the value of |z|.
\[Let\text{ }z\text{ }=\text{ }x\text{ }+\text{ }iy\] Now, \[\begin{array}{*{35}{l}} \Rightarrow ~{{x}^{2}}~\text{ }-1\text{ }+\text{ }{{y}^{2}}~=\text{ }0 \\ \Rightarrow ~{{x}^{2}}~+\text{...
Show that |(z – 2) / (z – 3)| = 2 represents a circle. Find its centre and radius.
\[\begin{array}{*{35}{l}} arg(\left| \left( z\text{ }\text{ }-2 \right)\text{ }/\text{ }\left( z-\text{ }\text{ }3 \right) \right|\text{ })=\text{ }2 \\ Substituting\text{ }z\text{ }=\text{...
If |z + 1| = z + 2 (1 + i), then find z.
\[\left| z\text{ }+\text{ }1 \right|\text{ }=\text{ }z\text{ }+\text{ }2\text{ }\left( 1\text{ }+\text{ }i \right)\] Substituting z = x + iy, we get, \[\Rightarrow ~\left| x\text{ }+\text{ }iy\text{...
Solve that equation |z| = z + 1 + 2i.
As per the inquiry, We have, \[\left| z \right|\text{ }=\text{ }z\text{ }+\text{ }1\text{ }+\text{ }2i\] Subbing z = x + iy, we get, \[\begin{array}{*{35}{l}} \Rightarrow \left| x\text{ }+\text{ }iy...
Show that the complex number z, satisfying the condition arg ((z-1)/(z+1)) = π/4 lies on a circle.
Let \[z\text{ }=\text{ }x\text{ }+\text{ }iy\] arg \[\left( \left( z-1 \right)/\left( z+1 \right) \right)\text{ }=\text{ }\pi /4\] \[\Rightarrow ~arg\text{ }\left( z\text{ }\text{ }1 \right)\text{...
If z = x + iy, then show that zz̅ + 2(z + z̅) + b = 0 where bϵR, representing z in the complex plane is a circle.
\[z\text{ }=\text{ }x\text{ }+\text{ }iy\] ⇒ z̅ = x – iy Now, we also have, z z̅ + 2 (z + z̅) + b = 0 \[\Rightarrow ~\left( x\text{ }+\text{ }iy \right)\text{ }\left( x\text{ }\text{ }iy...
If (1 + i)z = (1 – i) z̅, then show that z = i z̅.
SOLUTION: = -iz̅ Hence proved.
If a = cos θ + i sin θ, find the value of
SOLUTION: a = cos θ + i sin θ
IF
then find (a, b). Solution: = (i4)25 = 1 Hence, (a, b) = (1, 0)
If (FIG 1) then find the value of x + y.
FIG 1: SOLUTION: We have,
If (fig 1), then find (x, y).
fig 1: SOLUTION: We have,
Evaluate
where n ϵ N. SOLUTION: WE have,
For a positive integer n, find the value of (1 – i)^n (1 – 1/i)^n.
A/Q = (1 – i)n (1 + i)n = (1 – i2)n = 2n Therefore, (1 – i)n (1 – 1/i)n = 2n
If, then find the least positive integral value of m.
Solution:- Along these lines, the most un-positive number is \[1.\] Thus, the most un-positive necessary worth of \[m\text{ }is\text{ }4\text{ }\left( =\text{ }4\text{ }\times \text{ }1...
If
then show that
According to the given question, the solution should be
Find the number of non-zero integral solutions of the equation
Thus, \[0\]is the main necessary arrangement of the given condition. Thus, the quantity of non-zero necessary arrangements of the given condition is \[0.\]
If α and β are different complex numbers with |β| = 1, then find
Solution:- According to the given question, the solution should be
If
, then show that
Solution:- According to the given question, the solution should be
Find the modulus of
Solution:- According to the given question, the solution should be
Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.
Let’s suppose \[z\text{ }=\text{ }\left( x-iy \right)\text{ }\left( 3\text{ }+\text{ }5i \right)\] And \[\left( 3x\text{ }+\text{ }5y \right)\text{ }~i\left( 5x\text{ }-\text{ }3y \right)\text{...
Find the modulus and argument of the complex number
Solution:- According to the given question, the solution should be
Let
, find
(i) , (ii) Solution:- According to the given question, the solution should be
Solve:
Solution:- According to the given question, the solution should be
If
,Find
Solution:- According to the question, the solution should be Given,\[\text{ }{{z}_{1}}~=\text{ }2\text{ }-i,\text{ }{{z}_{2}}~=\text{ }1\text{ }+~i\]
Solve the following equation:
Given quadratic condition, \[21{{x}^{2}}-\text{ }28x~+\text{ }10\text{ }=\text{ }0\] On contrasting it and \[a{{x}^{2}}~+~bx~+~c~=\text{ }0,~\]we get \[a~=\text{ }21,~b~=-\text{ }28,\text{...
Solve the following equation:
Given quadratic condition, \[27{{x}^{2}}-\text{ }10x~+\text{ }1\text{ }=\text{ }0\] On contrasting it and\[a{{x}^{2}}~+~bx~+~c~=\text{ }0\], we get \[a~=\text{ }27,~b~=-\text{ }10,\text{...
Solve the following equation :
Given quadratic condition, \[{{x}^{2}}-\text{ }2x\text{ }+\text{ }3/2\text{ }=\text{ }0\] It very well may be re-composed as: \[~2{{x}^{2}}-\text{ }4x\text{ }+\text{ }3\text{ }=\text{ }0\] On...
Solve the following equation :
Given quadratic condition, \[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}-\text{ }\mathbf{4x}\text{ }+\text{ }\mathbf{20}/\mathbf{3}\text{ }=\text{ }\mathbf{0}\] It very well may be re-composed as:...
Convert the following in the polar form:
(I) (ii) Solution:- According to the given question, the solution should be,
Solve the following:
Solution:- According to the given question, the solution should be,
Reduce to the standard form
Solution:- According to the given question, the solution should be,
For any two complex numbers z1 and z2, prove that Re (z1z2) = Re z1 Re z2 – Im z1 Im z2
Solution:- According to the given question, the solution should be,
Evaluate:
Solution:- According to the given question, the solution should be
Solve the following equation:
Given quadratic condition, \[{{x}^{2}}~+~x/\surd 2~+\text{ }1\text{ }=\text{ }0\] It tends to be reworked as, \[\surd 2{{x}^{2}}~+~x~+\text{ }\surd 2\text{ }=\text{ }0\] On contrasting it and...
Solve the following equation:
Given quadratic condition, \[{{x}^{2}}~+~x~+\text{ }1/\surd 2\text{ }=\text{ }0\] It tends to be reworked as, \[\surd 2{{x}^{2}}~+\text{ }\surd 2x~+\text{ }1\text{ }=\text{ }0\] On contrasting it...
Solve the following equation:
Given quadratic condition, \[\surd 3{{x}^{2}}-\text{ }\surd 2x~+\text{ }3\surd 3\text{ }=\text{ }0\] On contrasting it and \[a{{x}^{2}}~+~bx~+~c~=\text{ }0,~\]we have \[a~=\text{ }\surd 3,~b~=\text{...
Solve the following equation:
Given quadratic condition, \[\surd 2{{x}^{2}}~+~x~+\text{ }\surd 2\text{ }=\text{ }0\] On contrasting it and\[a{{x}^{2}}~+~bx~+~c~=\text{ }0,\] we have \[a~=\text{ }\surd 2,~b~=\text{ }1,\text{...
Solve the following equation:
Given quadratic condition, \[{{x}^{2}}~-x~+\text{ }2\text{ }=\text{ }0\] On contrasting it and \[a{{x}^{2}}~+~bx~+~c~=\text{ }0,\]we have \[a~=\text{ }1,~b~=\text{ }1,\text{ }and~c~=\text{ }2\]...
Solve the following equation:
Given quadratic condition, \[{{x}^{2}}~+\text{ }3x~+\text{ }5\text{ }=\text{ }0\] On contrasting it and \[a{{x}^{2}}~+~bx~+~c~=\text{ }0,\]we have \[a~=\text{ }1,~b~=\text{ }3,\text{ }and~c~=\text{...
Solve the following equation:
Given quadratic condition, \[{{x}^{2}}~+~x-\text{ }2\text{ }=\text{ }0\] On contrasting it and\[a{{x}^{2}}~+~bx~+~c~=\text{ }0,~\]we have \[a~=\text{ }1,~b~=\text{ }1,\text{ }and~c~=\text{ }2\]...
Solve the following equation:
Given quadratic condition, \[{{x}^{2}}~+\text{ }3x~+\text{ }9\text{ }=\text{ }0\] On contrasting it and \[a{{x}^{2}}~+~bx~+~c~=\text{ }0,\]we have \[a~=\text{ }1,~b~=\text{ }3,\text{ }and~c~=\text{...
Solve the following equation:
Given quadratic condition, \[2{{x}^{2}}~+~x~+\text{ }1\text{ }=\text{ }0\] On contrasting it and\[a{{x}^{2}}~+~bx~+~c~=\text{ }0,\] we have \[a~=\text{ }2,~b~=\text{ }1,\text{ }and~c~=\text{ }1\]...
Solve the following equation:
Given quadratic condition, \[{{x}^{2}}~+\text{ }3\text{ }=\text{ }0\] On contrasting it and \[a{{x}^{2}}~+~bx~+~c~=\text{ }0,\] we have \[a~=\text{ }1,~b~=\text{ }0,\text{ }and~c~=\text{ }3\] Along...
Convert the complex numbers in the polar form i.
According to the given question, the solution should be
Convert the complex numbers in the polar form 3 + i
According to the given question, the solution should be
Convert the complex numbers in the polar form -3
According to the given question, the solution should be
Convert the complex numbers in the polar form – 1 – i
According to the given question, the solution should be
Convert the complex numbers in the polar form – 1 + i
According to the given question, the solution should be
Convert the complex numbers in the polar form 1 – i
According to the given question, the solution should be
Find the modulus and the arguments of the complex numbers z = -√3 + i
According to the given question, the solution should be
Find the modulus and the arguments of the complex numbers z = – 1 – i √3
According to the given question, the solution should be
Express the following expression in the form of a + ib:
SOLUTION:- According to the given question, the solution should be
Find the multiplicative inverse of the complex numbers – i
How about we think about \[z\text{ }=\text{ }\text{ }I\] Subsequently, the multiplicative backwards of \[\text{ }I\]is given by\[~z-1\]
Find the multiplicative inverse of the complex numbers √5 + 3i
How about we think about \[z\text{ }=\text{ }\surd 5\text{ }+\text{ }3i\] \[{{\left| z \right|}^{2}}~=\text{ }{{\left( \surd 5 \right)}^{2}}~+\text{ }{{3}^{2}}~=\text{ }5\text{ }+\text{ }9\text{...
Find the multiplicative inverse of the complex numbers 4 – 3i
How about we think about \[z\text{ }=\text{ }4\text{ }\text{ }3i\] Then, at that point, \[=\text{ }4\text{ }+\text{ }3i\] And \[{{\left| z \right|}^{2}}~=\text{ }{{4}^{2}}~+\text{ }{{\left( -3...
Express each of the complex number in the form a + ib. (-2 – 1/3i)3
Answer: According to question, the solution should be Hence, \[{{(-2\text{ }\text{ }1/3i)}^{3}}~=\text{ }-22/3\text{ }-\text{ }107/27i\]
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib. (1/3 + 3i)3
According to the question, the solution should be
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib. (1 – i)4
\[{{(1\text{ }~i)}^{4~}}=\text{ }{{[{{(1-\text{ }~i)}^{2}}]}^{2}}\] \[=\text{ }{{[1\text{ }+~{{i}^{2}}~-\text{ }2i]}^{2}}\] \[=\text{ }{{[1\text{ }-\text{ }1\text{ }-\text{...
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.
SOLUTION:- According to the question, the solution should be
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.
SOLUTION:- According to question, the solution should be:
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib. (1 – i) – (–1 + i6)
\[\begin{array}{*{35}{l}} \left( 1\text{ }-\text{ }i \right)-\text{ }\text{ }\left( \text{ }1\text{ }+\text{ }i6 \right)\text{ }=\text{ }1\text{ }-\text{ }i\text{ }+\text{ }1\text{ }-\text{ }i6 \\...
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib. 3(7 + i7) + i(7 + i7)
\[3(7\text{ }+~i7)\text{ }+~i(7\text{ }+~i7)~\] \[=\text{ }21\text{ }+~i21\text{ }+~i7\text{ }+~{{i}^{2~}}7\] \[=\text{ }21\text{ }+~i28\text{ }\text{ }7~\] \[[{{i}^{2}}~=\text{ }-1]\] \[=\text{...
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib. i-39
\[{{i}^{-39}}~=\text{ }1/\text{ }{{i}^{39}}~=\text{ }1/\text{ }{{i}^{4\text{ }x\text{ }9\text{ }+\text{ }3}}~\] \[=\text{ }1/\text{ }({{1}^{9}}~x\text{ }{{i}^{3}})\text{ }=\text{ }1/\text{...
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib. i9 + i19
\[{{i}^{9}}~+\text{ }{{i}^{19}}~=\text{ }{{({{i}^{2}})}^{4}}.\text{ }i\text{ }+\text{ }{{({{i}^{2}})}^{9}}.\text{ }i\] \[=\text{ }{{\left( -1 \right)}^{4}}~.\text{ }i\text{ }+\text{ }{{\left( -1...
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib. 1. (5i) (-3/5i)
$(5i)(-3/5i)=5x(-3/5)x{{i}^{2}}$ $=-3x-1[{{i}^{2}}=-1]$ \[=\text{ }3\] Consequently, \[\left( 5i \right)\text{ }\left( -\text{ }3/5i \right)\text{ }=\text{ }3\text{ }+\text{ }i0\]