Question #26c96

3 Answers
Jan 19, 2018

#-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#

Jan 19, 2018

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#

Jan 19, 2018

#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#