Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation z^6=-1
-
Equation cos(x/3)*cos(Pi/6)+sin(x/3)*sin(Pi/6)=-1
-
Equation ctgx=1
-
Equation sin^2((x/4)+(pi/4))*sin^2((x/4)-(pi/4))=0,375sin^2(-(pi/4))
- Express {x} in terms of y where:
- 5*x-5*y=20
- 8*x-7*y=-13
- 15*x-15*y=16
- 9*x+18*y=6
- Derivative of:
-
z^6
-
-1
- Graphing y =:
- z^6
- Integral of d{x}:
-
-1
- Limit of the function:
- -1
-
Identical expressions
- z^ six =- one
- z to the power of 6 equally minus 1
- z to the power of six equally minus one
- z6=-1
- z⁶=-1
-
Similar expressions
- log2(56)-(1/2log2(49))=x
- z^6=+1
z^6=-1 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$z^{6} = -1$$
Because equation degree is equal to = 6 and the free term = -1 < 0,
so the real solutions of the equation d'not exist
All other 6 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{6} = -1$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{6} e^{6 i p} = -1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{6 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = -1$$
so
$$\cos{\left(6 p \right)} = -1$$
and
$$\sin{\left(6 p \right)} = 0$$
then
$$p = \frac{\pi N}{3} + \frac{\pi}{6}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for w
Consequently, the solution will be for w:
$$w_{1} = - i$$
$$w_{2} = i$$
$$w_{3} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$w_{4} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$
$$w_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$w_{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - i$$
$$z_{2} = i$$
$$z_{3} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z_{4} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$
$$z_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z_{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$
$$z^{6} = -1$$
Because equation degree is equal to = 6 and the free term = -1 < 0,
so the real solutions of the equation d'not exist
All other 6 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{6} = -1$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{6} e^{6 i p} = -1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{6 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = -1$$
so
$$\cos{\left(6 p \right)} = -1$$
and
$$\sin{\left(6 p \right)} = 0$$
then
$$p = \frac{\pi N}{3} + \frac{\pi}{6}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for w
Consequently, the solution will be for w:
$$w_{1} = - i$$
$$w_{2} = i$$
$$w_{3} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$w_{4} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$
$$w_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$w_{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - i$$
$$z_{2} = i$$
$$z_{3} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z_{4} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$
$$z_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z_{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$
Sum and product of roots
[src]
sum
___ ___ ___ ___
I \/ 3 I \/ 3 \/ 3 I I \/ 3
-I + I + - - - ----- + - - ----- + ----- - - + - + -----
2 2 2 2 2 2 2 2
$$\left(- i\right) + \left(i\right) + \left(- \frac{\sqrt{3}}{2} - \frac{i}{2}\right) + \left(- \frac{\sqrt{3}}{2} + \frac{i}{2}\right) + \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right) + \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)$$
=
0
$$0$$
product
___ ___ ___ ___
I \/ 3 I \/ 3 \/ 3 I I \/ 3
-I * I * - - - ----- * - - ----- * ----- - - * - + -----
2 2 2 2 2 2 2 2
$$\left(- i\right) * \left(i\right) * \left(- \frac{\sqrt{3}}{2} - \frac{i}{2}\right) * \left(- \frac{\sqrt{3}}{2} + \frac{i}{2}\right) * \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right) * \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)$$
=
1
$$1$$
Rapid solution
[src]
z_1 = -I
$$z_{1} = - i$$
z_2 = I
$$z_{2} = i$$
___
I \/ 3
z_3 = - - - -----
2 2
$$z_{3} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
___
I \/ 3
z_4 = - - -----
2 2
$$z_{4} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$
___
\/ 3 I
z_5 = ----- - -
2 2
$$z_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
___
I \/ 3
z_6 = - + -----
2 2
$$z_{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$
Numerical answer
[src]
z1 = 0.866025403784439 + 0.5*i
z2 = -0.866025403784439 - 0.5*i
z3 = -0.866025403784439 + 0.5*i
z4 = 1.0*i
z5 = 0.866025403784439 - 0.5*i
z6 = -1.0*i
z6 = -1.0*i
The graph