Tuesday

July 29, 2014

July 29, 2014

Posted by **Anonymous** on Monday, March 12, 2007 at 9:25am.

i got up to z^3 = 4+3i and 4-3i then got stuck !

4 + 3i = 5 Exp[i theta]

The equation z^3 = Q for real positive Q has three solutions:

z = cuberoot[Q] Exp[2 pi n i/3]

for n = 0, 1 and 2, because

Exp[2 pi n i/3]^3 = Exp[2 pi n i] = 1

So, in this case you find:

z = 5^(1/3) Exp[i theta/3 + 2 pi n i/3 ]

**Related Questions**

Abstract Algebra - Let H={a+bi a,b is a element R, a^2+b^2=1} be a subset of the...

Mathematics - Express the Complex Number -1-i in polar form. how is complex no ...

complex numbers - -3 + 4i in polar form in pi the magnitude is 5 from pyth ...

math-complex numbers - Sketch the sets of complex numbers in the complex plane ...

Math (Complex Numbers) - Let a,b,c be complex numbers satisfying a+b+c=abc=1 and...

calculus - can you please explain tep by step how to solve dividing complex ...

complex numbers! - -2-2i in polar form = 2.8 cis 2.35?? -3+4i polar form = 5 cis...

math - solve and simplify the answer completely by using pure imaginary numbers/...

Math - I am in the 5th grade and my homework tonight is about complex numbers. I...

trig - Find z1z2 and z1/z2 for the pair of complex numbers using trigonometric ...