Question

Question #26c96

Answer 1

`-3i+1`

Explanation:

You have `((2-i)/(1-i))*(-2i)`.
Do the multiplication:

`((2-i)*(-2i))/(1-i)`

`(-4i-2)/(1-i)`

Rationalizing the denominator:

`(-4i-2)/(1-i)*(1+i)/(1+i)`

`((1+i)*(-4i-2))/2`

`((-4i-2)+i(-4i-2))/2`

`(-4i-2+4-2i)/2`

`(-6i+2)/2`

`-3i+1`

Answer 2

Multiply both numerator and denominator by the complex conjugate of the denominator to obtain:

`(2-i)/(1-i) * (-2i) = 1-3i`

Explanation:

The trick for dividing complex numbers is to multiply both numerator and denominator by the complex conjugate of the denominator.

`(2-i)/(1-i) * (-2i)`

`-> (2-i)/(1-i) * (1+i)/(1+i) * (-2i)`

`-> (2+2i-i+1)/(1+i-i+1)*(-2i)`

`-> ((3+i)* (-2i))/(2)`

`-> 1-3i`

Answer 3

`1-3i`

Explanation:

Some things to consider.

`i = sqrt(-1)`

`i^2=(sqrt(-1))^2=-1`

This idea extends to higher powers.

`i^3=(sqrt(-1))^3=-sqrt(-1)=-i`

`i^4=(sqrt(-1))^4= (-1)(-1)=1` etc

We will not need powers greater than 2 for this problem.

Also note, for any complex number:

`a+bi`

Where a and b are real numbers and `i=sqrt(-1)`

There exist a complex number `a-bi` , such that:

`(a+bi)*(a-bi)=a^2+b^2`

Which is always a real number.

This is known as a complex conjugate.

For the example:

`(2-i)/(1-i)*(-2i)`

We will start with `(2-i)/(1-i)`

Dividing by a complex number is undefined, so we have to remove the complex number from the denominator by multiplying by the complex conjugate of the denominator:

`((1+i)(2-i))/((1+i)(1-i))`

Multiplication is carried out in the normal way, but we have to be aware when we multiply `ixxi`, which we know is `-1`

`((1+i)(2-i))/((1+i)(1-i))=((1*2)-(i*1)+(2*i)-(i*i))/((1*1)-(1*i)+(1*i)-(i*i))`

`=((2)-(i)+(2i)-(-1))/((1)-(1i)+(1i)-(-1))=(3+i)/(2)`

Notice `color(blue)(i +2i = 3i)`. Apart from when we multiply i by itself it is treated just like a variable, such as `x`

So we have so far:

`(3+i)/(2)`

Now we multiply by `-2i`

`((3+i) * (-2i)) /2=((3) * (-2i )+(i*(-2i)))/2=((-6i )+(2))/2=1-3i`