Question 1196157
.
Find the modulus and argument of √(1+2i)/(1-2i)
~~~~~~~~~~~~~~~~~



            The solution by  Edwin is fine.

            Another solution and another analysis is possible,  which is in my post below.


            It allows to get the answer with significantly less computational efforts.



<pre>
So, we consider  {{{sqrt((1+2i)/(1-2i))}}}.    (1)


First consider the ratio under the square root  {{{(1+2i)/(1-2i)}}}.    (2)


The numerator has the modulus  |1+2i| = {{{sqrt(1^2 + 2^2)}}} = {{{sqrt(5)}}}.

The denominator has the modulus  |1-2i| = {{{sqrt(1^2 + (-2)^2)}}} = {{{sqrt(5)}}}.


Therefore, the ratio  {{{(1+2i)/(1-2i)}}}  has the modulus of  {{{sqrt(5)/sqrt(5)}}} = 1.


Hence, the original complex number {{{sqrt((1+2i)/(1-2i))}}}  has the modulus  {{{sqrt(1)}}} = 1.


    +-----------------------------------------------+
    |    Thus we completed with the modulus.        |
    |    Now, let's determine the argument of (1).  |
    +-----------------------------------------------+



Consider the ratio under the square root {{{highlight(highlight(again))}}}  {{{(1+2i)/(1-2i)}}}.    (3)


The numerator has the argument "a" in QI such that  tan(a) = 2/1 = 2.

The denominator has the argument "b" in QIV such that tan(b) = (-2)/1 = -2.

Therefore, b = -a  (which is obvious).

Hence, the ratio  {{{(1+2i)/(1-2i)}}}  has the argument  a - b = a - (-a) = 2a.


When we take square root of (3), the argument of the resulting complex number will be {{{(2a)/2}}} = a.


Thus we get the <U>ANSWER</U>: 

    +----------------------------------------------------------------------------+
    |    the modulus of the sough complex number is 1;                           |
    |                                                                            |
    |    the argument of the sought complex number is "a" such that tan(a) = 2,  |
    |        i.e. a = arctan(2) = 1.10715 radians = 63.435 degrees (rounded).    |     
    +----------------------------------------------------------------------------+


For completeness, when we take square root of (3), we obtain, actually, two complex numbers.
First one has the argument "a", as we found it above; the second complex number has the argument a+{{{pi}}}.

So, the second value of the square root has the modulus of 1 and the argument arctan(2)+{{{pi}}} = 243.435 degrees.
</pre>

Solved.


The answer is the same as in the post by Edwin.


-----------------


On complex numbers, &nbsp;see relevant lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/Complex-numbers-and-arithmetical-operations.lesson>Complex numbers and arithmetical operations on them</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/Complex-plane.lesson>Complex plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/Addition-and-subtraction-of-complex-numbers-in-complex-plane.lesson>Addition and subtraction of complex numbers in complex plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/Multiplication-and-division-of-complex-numbers-in-complex-plane-.lesson>Multiplication and division of complex numbers in complex plane</A>


&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/Solved-problems-on-taking-roots-of-complex-numbers.lesson>Solved problems on taking roots of complex numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/Solved-problems-on-arithmetic-operations-on-complex-numbers.lesson>Solved problems on arithmetic operations on complex numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/Solved-problem-on-taking-square-roots-of-complex-numbers.lesson>Solved problem on taking square root of complex number</A>

in this site.


Also, &nbsp;you have this free of charge online textbook on &nbsp;ALGEBRA-II &nbsp;in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic &nbsp;"<U>Complex numbers</U>".



Save the link to this textbook together with its description


Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson


into your archive and use when it is needed.