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Question

Question: Simplify \(\dfrac{3-2i}{-4-i}\) ?...

Simplify 32i4i\dfrac{3-2i}{-4-i} ?

Explanation

Solution

Here we know that the given above equation is the form of division it basically form of complex numbers, where in the above equation we have to multiply both the numerator as well as the denominator with its conjugates to solve it further.

Complete step by step solution:
In the given numerical we know I is the form of complex number but the form of division, where we have to multiply the equation with its conjugates,
So we have to use the formula of division are as follow,
a+bic+di=(a+bi)(cdi)(c+di)(cdi)\dfrac{a+bi}{c+di}=\dfrac{\left( a+bi \right)\left( c-di \right)}{\left( c+di \right)\left( c-di \right)}
As we know that we have to multiply the equation with its conjugates,
Thus substituting the equation in the division formula we get,
32i4i=(32i)(4+i)(4i)(4+i)\dfrac{3-2i}{-4-i}=\dfrac{\left( 3-2i \right)\left( -4+i \right)}{\left( -4-i \right)\left( -4+i \right)}
So in the above equation we have multiply with its (4+i)\left( -4+i \right) both with numerator as well as denominator,
Further opening the bracket we get, multiplying (32i)(4+i)\left( 3-2i \right)\left( -4+i \right)
So we have to multiply with three into four, again three with i,2i-2i with 4-4 and 2i-2i with I we get,
=12+3i+8i2i2(4i)(4+i)=\dfrac{-12+3i+8i-2{{i}^{2}}}{\left( -4-i \right)\left( -4+i \right)}
Again opening the lower bracket we have to multiply (4i)(4+i)\left( -4-i \right)\left( -4+i \right), here we are multiplying 4-4 with 4-4 , 4-4 with i,ii,-i with 4-4 and ii with +i+i thus we obtained.
12+3i+8i2i2164i+4ii2\dfrac{-12+3i+8i-2{{i}^{2}}}{16-4i+4i-{{i}^{2}}}
Here we will add the value of 3i+8i3i+8i were we are going to obtained as 11i11i also when we are adding 4i4i with +4i+4i it will going to cancel it each other and remaining will be zero.
=12+11i2i216i2=\dfrac{-12+11i-2{{i}^{2}}}{16-{{i}^{2}}}
We know that the value of i2=1{{i}^{2}}=-1 so substituting the value in above equation we get,
=12+11i2(1)16(1)=\dfrac{-12+11i-2\left( -1 \right)}{16-\left( -1 \right)}
So further solving the above equation we get as
=10+11i17=\dfrac{-10+11i}{17}
Hence while solving the above numerical we get the value as =10+11i17=\dfrac{-10+11i}{17}

Note: First we had to identify the equation of complex number then we have to look its division farm, Applying the division rule, solve equation using BODMAS rule,
Remember the value of i2=1{{i}^{2}}=1