Solving z^4=16i: Finding All Possible Solutions

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Homework Help Overview

The problem involves finding all complex numbers \( z \) such that \( z^4 = 16i \). The original poster seeks solutions in rectangular form without using trigonometric functions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to express \( z \) in exponential form and questions the correct placement of the \( 2n\pi \) term in relation to taking the fourth root. They also inquire about which values of \( n \) to use to find the four solutions.
  • Some participants clarify the placement of the \( 2n\pi \) term and suggest using consecutive integers for \( n \) to find the roots.
  • There is a mention of needing trigonometric functions for converting the results to rectangular form, which raises questions about the original poster's constraints.

Discussion Status

The discussion is active, with participants providing clarifications on the placement of terms and the selection of \( n \). There is no explicit consensus yet, as participants are exploring different aspects of the problem.

Contextual Notes

The original poster has specified a preference for avoiding trigonometric functions in their solutions, which may influence the discussion on how to express the roots.

Physics_wiz
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Need to find all z such that z^4 = 16i. Rectangular form and no trig functions. Here's what I did:
z^4 = 16e^(i pi/2) = 16e^i(pi/2 + 2npi)

z = 2e^i(pi/8 + npi/2)

First question: Do I add a 2npi before I take the 4th root or do I add it after I take the 4th root to get z = 2e^i(pi/8 + 2npi)? Does it matter?

Second question: After I get an expression for z, which n's do I plug in the equation to find the 4 z's I'm looking for? How do I know that?
 
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First question : BEFORE
second question : take n : 0,1,2,3

marlon
 
G'day, Physics_wiz.

Physics_wiz said:
First question: Do I add a 2npi before I take the 4th root or do I add it after I take the 4th root to get z = 2e^i(pi/8 + 2npi)? Does it matter?
It must be before. You might write it as
Let
{z_n}^4 = 16e^{i\left(\frac{\pi}{2} + 2n\pi \right)}

So
z_n = 2e^{i\left(\frac{\frac{\pi}{2} + 2n\pi}{4}\right)} = 2e^{i\left(\frac{\pi}{8} + \frac{n\pi}{2}\right)}

The four complex roots will be an angle of \frac{2\pi}{4} apart on an Argand diagram.

Physics_wiz said:
Second question: After I get an expression for z, which n's do I plug in the equation to find the 4 z's I'm looking for? How do I know that?
Any consecutive four; that is how many complex fourth roots we expect of z^4. eg. n=0, 1, 2, 3; or n=4, 5, 6, 7; etc. The convention is to choose consecutive values for n such that the argument is no greater than pi, though.

Note that you will need the trig functions on your calculator to convert the roots you find in exponential form to rectangular form, so I'm not exactly sure what the question means by "no trig functions".
 
Last edited:
Unco said:
{z_n}^4 = 16e^{\left(\frac{\pi}{2}i+ 2n\pi \right)}
No, it must be :
{z_n}^4 = 16e^{ i \left (\frac{\pi}{2} + 2n\pi \right)}
marlon
 
Last edited:
Thank you for the correction. Is there a way to preview the Latex output?
 
Unco,

Sorry, not right now. Look at this message

regards
marlon
 

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