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What is the sum of the serious if n equals 5 and ai equals 4i plus 3?
I cannot take this question seriously! The sum is 75.
What are the cube roots of 1000i?
5√3 + 5i, -5√3 + 5i, -10i
What is the multiplicative inverse of 3-4i divided by 5 plus 2i?
The multiplicative inverse of a number a is a number b such that axb=1 If we look at (3-4i)/(5+2i), we see that we can multiply that by its reciprocal and the product is one. So (5+2i)/(3-4i) is the multiplicative inverse of (3-4i)/(5+2i)
What is X when raised to the 3-5i power and the answer is 23-14i?
X = 1.31356+0.612045*iSteps to solve, take the natural log of both sides:ln(X^(3-5i)) = ln(23-14i).(3-5i)*ln(X) = ln(23-14i). Convert 23-14i to exponential form: A*e^(iΘ) {A = 26.926 and Θ = -0.54679 radians}(3-5i)*ln(X) = ln(A*e^(iΘ))= ln(A) + iΘ = ln(26.926) - 0.54679i.divide by (3-5i): ln(X) = (ln(A) + iΘ) / (3-5i) = (3.2931 - 0.54679i)/(3-5i)So we have ln(X) = 0.370978 + 0.436033i, then:e^(ln(X)) = e^(0.370978 + 0.436033i) --> X = 1.31356+0.612045*i
What is the sum of 5 3 - - 12 and 12?
It is: 5 + 3 - - 12+12 = 32
Plot the number in a complex plane -1-3i?
2
What is the reciprocal of 3 plus 5i?
1/(3+5i)=(3-5i)/((3+5i)(3-5i))=(3-5i)/(9+25)=(3-5i)/34
What is the sum of the serious if n equals 5 and ai equals 4i plus 3?
I cannot take this question seriously! The sum is 75.
How do you find the additive inverse of a complex number?
You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).
Complex conjugate of 5i?
0 + 5i Its complex conjugate is 0 - 5i
What are 10 complex numbers with the absolute value of 5?
Use the Pythagorean theorem. 5, -5, 5i, and -5i will work, as well as any combination of a real and imaginary number such that (real part) squared + (imaginary part) squared = 25, for example, 4 + 3i, 3 + 4i, 4 - 3i, etc.
What is the conjugate of 3 plus 4i?
When finding the conjugate of a binomial, you just reverse the sign. So the conjugate of 3+4i is 3-4i.
What are the cube roots of 1000i?
5√3 + 5i, -5√3 + 5i, -10i
Convert to polar form and then divide Express your answer in polar form 5 sqare root of 3 -5i divided by 4 plus 4isquare root of 3?
(5√3 - 5i)/(4 + 4i√3)Let's try to represent the given complex numbers in the polar form.z = |z|(cos θ + i sin θ)Let z1 = 5√3 - 5i in the form a + bi, wherea1 = |z1|cos θ1 andb1 = |z1| sin θ1so that|z1| = √(a12 + b12) = √[(5√3)2 + (- 5)2] = √(75 + 25) = √100 = 10cos θ = a1/|z1|= 5√3/10 = √3/2 andsin θ = b1/|z1|= -5/10 = -1/2So that,z1=5√3 - 5=|z1|(cos θ1 + i sin θ1) =10(√3/2 - i1/2), where θ1 = 330 degreesLet z2 = 4 + 4√3i|z2|=√(42 + (4√3)2) = √(16 + 48) = √64= 8cos θ2 = 4/8 = 1/2sin θ2 = (4√3)/8 = √3/2So that,z2 = 4 + 4√3 = |z2|(cos θ2 + i sin θ2) = 8(1/2 + √3/2i), where θ2 = 60 degreesNow let's divide.Recall the Euler formula: z = |z|eiθ, wher eiθ = (cos θ + i sin θ)z1/z2 = |z1|eiθ1/|z2|eiθ2 = (|z|/|z|)ei(θ1 - θ2) = (|z|/|z|)[cos (θ1 - θ2) + i sin (θ1 - θ2)]z1/z2 = (10/8)[cos (330 - 60) + i sin (330 - 60)] = (5/2)( cos 270 + i sin 270) = (5/4)(0 - i )Let's check:(5√3 - 5i)/(4 + 4i√3)= (5√3 - 5i)(4 - 4i√3)/(4 + 4i√3)(4 - 4i√3)= (20√3 - 60i - 20i + 20i2√3)/(16 - 48i2)= (20√3 - 80i - 20√3)/(16 + 48)= -80i/64= -5i/4= 0 - 5i/4= (5/4)(0 - i)
What is the sum of -3 and -12?
9
What is the multiplicative inverse of 3-4i divided by 5 plus 2i?
The multiplicative inverse of a number a is a number b such that axb=1 If we look at (3-4i)/(5+2i), we see that we can multiply that by its reciprocal and the product is one. So (5+2i)/(3-4i) is the multiplicative inverse of (3-4i)/(5+2i)
What is the product of these complex numbers (3-4i)(1-i)?
(3-4i)(1-i) = (3x1) + (3 x -i) + (-4i x 1) + ( -4i x -i) = 3 - 3i -4i -4 = -1 - 7i
