Complex number

[Erros]

How to do this task I have no idea or idea how

 

[topsquark]

Let z = a + ib. Then
$\dfrac{z}{z_1} = \dfrac{a + ib}{2 + i}$ Hint: Multiply top and bottom by 2 - i.

$\overline{z} z_1 = (a - ib)(2 + i)$. Hint: Multiply this out.

You will get two equations in a and b to solve simultaneously.

Let us know what you are able to do.

-Dan

 

[Erros]

But why are the results of Re and Im given?

Let z = a + ib. Then
$\dfrac{z}{z_1} = \dfrac{a + ib}{2 + i}$ Hint: Multiply top and bottom by 2 - i.

$\overline{z} z_1 = (a - ib)(2 + i)$. Hint: Multiply this out.

You will get two equations in a and b to solve simultaneously.

Let us know what you are able to do.

-Dan

 

[Harry_the_cat]

But why are the results of Re and Im given?

So that you have something to equate your expressions involving a and/or b to.

 

[topsquark]

But why are the results of Re and Im given?

They are just part of the equations that gives a and b. It's like asking for the integer part of a number or telling you to choose the negative value of the square root in an equation.

-Dan

 

[pka]

How to do this task I have no idea or idea how

Here with some change jn notation is an approach.
If $z=a+bi,~w=2+i,~\Re\left(\dfrac{z}{w}\right)=\dfrac{-3}{5},~\&~\Im\left(\overline{~z~}\cdot w\right)=1$.
You should know tht $\dfrac{1}{w}=\dfrac{\overline{~w~}}{|w|^2}$ Thus $\left(\dfrac{z}{w}\right)=\dfrac{z\cdot\overline{~w~}}{5}$
Then $(a+bi)(2-i))=(2a+b)+i(2b-a)$ Now use the given the finish an post your findings.


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