Complex number q (1 Viewer)

[Joshmosh2]

How do you work out the square root of 12-5i ?

By doing sqrt (12-5i) = a+ib and solving simultaneously, he result is a crazy quartic eqn which can be difficult to find the roots.

What is the simplest way of approaching this q? Thanks.

 

[trapizi]

Do what he teaches you.
sqrt(12-5i)=a+bi
12-5i=(a+bi)^2
Re: x^2-y^2=12
Im xy=-5/2
Let y= -5/2x or x=-5/2y and sub it in the Re equation.
Times both side by x^2 if you make a y sub or y^2 if you make an x sub
Solve it like quadratic equation and then square root the real positive root (usually).
There's another way but requires you to remember long equation so it's just make the problem more complicated.

 

[faisalabdul16]

hahaha second day of 4u, already posting questions

 

[Drongoski]

Or you can use the method used in Terry Lee's:

12-5i = 0.5(24-10i) = 0.5(25-10i-1)= 0.5[5^2 + 2x5x(-i) + (-i)^2] = 0.5[(5-i}^2]

 

[seventhroot]

hahaha second day of 4u, already posting questions

ikr lel

 

[Joshmosh2]

hahaha second day of 4u, already posting questions

Weirdly enough, i'm starting next week o.o

 

[Joshmosh2]

Or you can use the method used in Terry Lee's:

12-5i = 0.5(24-10i) = 0.5(25-10i-1)= 0.5[5^2 + 2x5x(-i) + (-i)^2] = 0.5[(5-i}^2]

I tried to understand this method, but I'm not good at manipulating equations into fractions like that

 

[Joshmosh2]

Or you can use the method used in Terry Lee's:

12-5i = 0.5(24-10i) = 0.5(25-10i-1)= 0.5[5^2 + 2x5x(-i) + (-i)^2] = 0.5[(5-i}^2]

Can you please clarify this method to me? what's the purpose of doing this?
So you manipulate the terms inside to bracket to safely remove the square root?

 

[faisalabdul16]

This is how you can do it

 

[Joshmosh2]

One more thing,
Given that z = x+yi, where x, y are real numbers. If (iz-1)/(z-i) is a real number, show that when (x,y) =/ (0,1), x^2 + y^2 =1

 

[turntaker]

You can use eulers form to express 12-5i in mod-arg form. Then make that to the power of 1/2 which is the same thing as sqrt().
Idk if this helped but its a REALLY cool method

 

[Drongoski]

Can you please clarify this method to me? what's the purpose of doing this?
So you manipulate the terms inside to bracket to safely remove the square root?

Idea is to convert, through appropriate algebraic fiddling to convert your complex number into a perfect square - i.e. (xxx)². Then the square root of your given complex number is simply + or - "xxx".

 

[glittergal96]

One more thing,
Given that z = x+yi, where x, y are real numbers. If (iz-1)/(z-i) is a real number, show that when (x,y) =/ (0,1), x^2 + y^2 =1

Multiply numerator and denominator by the conjugate of the denominator. The numerator of this new fraction must then be real. This gives us



The imaginary part of the LHS is then



which must be equal to zero.