How to Solve Complex Number Equations with Modulus and Argument?

[HMPARTICLE]

Homework Statement

a) Find the modulus and argument of 6^(1/2) + 2^(1/2)i

b) Solve the equation z^(3/4) = 6^(1/2) + 2^(1/2)i

Homework Equations

The Attempt at a Solution

For part a) i used Pythagoras to find the modulus.

( (6^(1/2))^2 + (2^(1/2))^2 )^(1/2) = (6 + 2)^(1/2) = 8^(1/2)

and to find the principle argument of z i said arctan(2^(1/2)/6^(1/2) = pi/6

I assume this is correct.

for part b)

z^(3/4) = 6^(1/2) + 2^(1/2)i

Then

z^(3/4) = 8^(1/2)(cos(pi/6 + 2kpi) +isin(pi/6 + 2kpi))

Then raising the R.H.S of the Equation to 4/3 to get z and Then using de Moivre's theorem.

therefore z = 4(cos(4pi/18 + 8kpi/3) +isin(4pi/18 + 8kpi/3))


This is where i am stuck. As k \in Z i can get a value of theta which is contained within
-pi < theta <= pi when k = 0.
if i assign a value of 1 to K then i get a value of \vartheta which is greater than pi.
the answer i get when K = 0 is wrong and there are 3 solutions. So i must be doing something wrong elsewhere.

I'm sorry in advance if i have made a silly error.

 

[nrqed]

Homework Statement

a) Find the modulus and argument of 6^(1/2) + 2^(1/2)i

b) Solve the equation z^(3/4) = 6^(1/2) + 2^(1/2)i

Homework Equations

The Attempt at a Solution

For part a) i used Pythagoras to find the modulus.

( (6^(1/2))^2 + (2^(1/2))^2 )^(1/2) = (6 + 2)^(1/2) = 8^(1/2)

and to find the principle argument of z i said arctan(2^(1/2)/6^(1/2) = pi/6

I assume this is correct.

for part b)

z^(3/4) = 6^(1/2) + 2^(1/2)i

Then

z^(3/4) = 8^(1/2)(cos(pi/6 + 2kpi) +isin(pi/6 + 2kpi))

Then raising the R.H.S of the Equation to 4/3 to get z and Then using de Moivre's theorem.

therefore z = 4(cos(4pi/18 + 8kpi/3) +isin(4pi/18 + 8kpi/3))


This is where i am stuck. As k \in Z i can get a value of theta which is contained within
-pi < theta <= pi when k = 0.
if i assign a value of 1 to K then i get a value of \vartheta which is greater than pi.
the answer i get when K = 0 is wrong and there are 3 solutions. So i must be doing something wrong elsewhere.

I'm sorry in advance if i have made a silly error.

Hi!

I want to focus first on your result for k=0. It looks right to me. What value are you supposed to get ? (this may be a silly comment but if you calculated a numerical value, were you in radians?)

 

[HMPARTICLE]

The answer 7b are the values I should be getting :/

 

[ehild]

z^(3/4) = 8^(1/2)(cos(pi/6 + 2kpi) +isin(pi/6 + 2kpi))

Then raising the R.H.S of the Equation to 4/3 to get z and Then using de Moivre's theorem.

therefore z = 4(cos(4pi/18 + 8kpi/3) +isin(4pi/18 + 8kpi/3))


This is where i am stuck. As k \in Z i can get a value of theta which is contained within
-pi < theta <= pi when k = 0.
if i assign a value of 1 to K then i get a value of \vartheta which is greater than pi.
the answer i get when K = 0 is wrong and there are 3 solutions. So i must be doing something wrong elsewhere.

I'm sorry in advance if i have made a silly error.

First rise the equation to power 4. You get z³. Then take the third roots of the number with k=0.