Question

Determine the complex number z which satisfies z (3 + 3i) = 2 – i.

Answer 1

Hint: First of all separate z by dividing both sides by 3 + 3i. Now divide and multiply RHS by (3 – 3i) and use the formula \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\] to get the value of z.

Complete step-by-step answer:
In this question, we have to find the value of z which satisfies z (3 + 3i) = 2 – i. Let us consider the equation given in the question
\[z\left( 3+3i \right)=2i\]
First of all, let us separate z from another term of the above equation. So, by dividing (3 + 3i) on both the sides of the above equation, we get,
\[\dfrac{z\left( 3+3i \right)}{\left( 3+3i \right)}=\dfrac{\left( 2-i \right)}{\left( 3+3i \right)}\]
By canceling the like terms of the above equation, we get,
\[z=\dfrac{\left( 2-i \right)}{\left( 3+3i \right)}\]
Now, by multiplying (3 – 3i) on both the denominator and numerator of the right-hand side (RHS) of the above equation, we get,
\[z=\dfrac{\left( 2-i \right)}{\left( 3+3i \right)}\times \dfrac{\left( 3-3i \right)}{\left( 3-3i \right)}\]
\[\Rightarrow z=\dfrac{\left( 2-i \right)\left( 3-3i \right)}{\left( 3+3i \right)\left( 3-3i \right)}\]
We know that \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]. By considering a = 3 and b = 3i, we get,
\[z=\dfrac{\left( 2-i \right)\left( 3-3i \right)}{\left[ {{\left( 3 \right)}^{2}}-{{\left( 3 \right)}^{2}}{{\left( i \right)}^{2}} \right]}\]
By simplifying the above equation, we get,
\[z=\dfrac{\left( 2 \right)\left( 3 \right)-2\left( 3i \right)-i\left( 3 \right)+3{{\left( i \right)}^{2}}}{\left[ {{\left( 3 \right)}^{2}}-{{\left( 3 \right)}^{2}}{{\left( i \right)}^{2}} \right]}\]
We know that, \[i=\sqrt{-1}\], so we get \[{{i}^{2}}=-1\]. So, by substituting \[{{i}^{2}}=-1\] in the above equation, we get,
\[z=\dfrac{6-6i-3i+3\left( -1 \right)}{\left( 9+9 \right)}\]
\[z=\dfrac{6-9i-3}{18}\]
\[\Rightarrow z=\dfrac{3-9i}{18}\]
We can also write the above equation as,
\[\Rightarrow z=\dfrac{3\left( 1-3i \right)}{18}\]
\[\Rightarrow z=\dfrac{\left( 1-3i \right)}{6}=\dfrac{1}{6}-\dfrac{i}{2}\]
Hence, we get the value of z as \[\dfrac{\left( 1-3i \right)}{6}\] or \[\dfrac{1}{6}-\dfrac{i}{2}\].

Note: In this question, students can cross-check their solution as follows:
The equation given in the question is, z (3 + 3i) = 2 – i
By substituting \[z=\dfrac{\left( 1-3i \right)}{6}\], we get,
\[\Rightarrow \dfrac{\left( 1-3i \right)}{6}\left( 3+3i \right)=\left( 2-i \right)\]
\[\Rightarrow \dfrac{3\left( 1-3i \right)\left( 1+i \right)}{6}=\left( 2-i \right)\]
\[\Rightarrow \dfrac{\left( 1-3i \right)\left( 1+i \right)}{2}=\left( 2-i \right)\]
\[\Rightarrow \dfrac{1+i-3i-3{{\left( i \right)}^{2}}}{2}=\left( 2-i \right)\]
\[\Rightarrow \dfrac{1-2i+3}{2}=\left( 2-i \right)\]
\[\Rightarrow \dfrac{4-2i}{2}=\left( 2-i \right)\]
\[\Rightarrow \left( 2-i \right)=\left( 2-i \right)\]
LHS = RHS
Here, we get LHS = RHS, hence our value of z is correct.