How do you simplify #-2/(3-i)#?

2 Answers
Dec 29, 2015

Answer:

Multiply the numerator and denominator by the conjugate of the denominator to find

#-2/(3-i) = -3/5 - 1/5i#

Explanation:

The conjugate of a complex number #a+bi# is #a-bi#. The product of a complex number and its conjugate is a real number. We will use this property to produce a real number in the denominator of the given expression.

#-2/(3-i) = -2/(3-i)*(3+i)/(3+i)#

#= (-2(3+i))/((3-i)(3+i))#

#= (-6-2i)/(9 +3i -3i +1)#

#= (-6-2i)/10#

#= -3/5 - 1/5i#

Dec 29, 2015

Answer:

#=0.632/_0.3217#. (in polar form)

#=-0.6-0.2i#. (in rectangular form)

Explanation:

There are 2 methods to do this -
Method 1
First convert everything to polar form then use the formula
#(z_1)/(z_2)=(r_1)/(r_2)cis(theta_1-theta_2)#

#therefore (-2/pi)/((sqrt(3^2+1^2))/_tan^-1(-1/3))=-2/sqrt10 /_ (pi-(-0.32175)#

#=0.632/_3.4633#. (Principle angle arguement is 0.3217 rad).

Method 2
Multiply the quantity by 1, selecting 1 as the complex conjugate of the denominator over itself.
Then the resultant denominator is a real number, and the numerator we multiply in rectangular form using the rule
#(a+ib)*(x+iy)=(ax-by)+i(bx+ay)#.

#therefore (-2+0i)/(3-i)*(3+i)/(3+i)=(-6-2i)/(3^2+1^2)#

#=1/10(-6-2i)#

#=-0.6-0.2i#