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

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

Explanation

Solution

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 (a+b)(a−b)=a2−b2\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(3+3i)=2iz\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,
z(3+3i)(3+3i)=(2−i)(3+3i)\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=(2−i)(3+3i)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=(2−i)(3+3i)×(3−3i)(3−3i)z=\dfrac{\left( 2-i \right)}{\left( 3+3i \right)}\times \dfrac{\left( 3-3i \right)}{\left( 3-3i \right)}
⇒z=(2−i)(3−3i)(3+3i)(3−3i)\Rightarrow z=\dfrac{\left( 2-i \right)\left( 3-3i \right)}{\left( 3+3i \right)\left( 3-3i \right)}
We know that (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}. By considering a = 3 and b = 3i, we get,
z=(2−i)(3−3i)[(3)2−(3)2(i)2]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=(2)(3)−2(3i)−i(3)+3(i)2[(3)2−(3)2(i)2]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=−1i=\sqrt{-1}, so we get i2=−1{{i}^{2}}=-1. So, by substituting i2=−1{{i}^{2}}=-1 in the above equation, we get,
z=6−6i−3i+3(−1)(9+9)z=\dfrac{6-6i-3i+3\left( -1 \right)}{\left( 9+9 \right)}
z=6−9i−318z=\dfrac{6-9i-3}{18}
⇒z=3−9i18\Rightarrow z=\dfrac{3-9i}{18}
We can also write the above equation as,
⇒z=3(1−3i)18\Rightarrow z=\dfrac{3\left( 1-3i \right)}{18}
⇒z=(1−3i)6=16−i2\Rightarrow z=\dfrac{\left( 1-3i \right)}{6}=\dfrac{1}{6}-\dfrac{i}{2}
Hence, we get the value of z as (1−3i)6\dfrac{\left( 1-3i \right)}{6} or 16−i2\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=(1−3i)6z=\dfrac{\left( 1-3i \right)}{6}, we get,
⇒(1−3i)6(3+3i)=(2−i)\Rightarrow \dfrac{\left( 1-3i \right)}{6}\left( 3+3i \right)=\left( 2-i \right)
⇒3(1−3i)(1+i)6=(2−i)\Rightarrow \dfrac{3\left( 1-3i \right)\left( 1+i \right)}{6}=\left( 2-i \right)
⇒(1−3i)(1+i)2=(2−i)\Rightarrow \dfrac{\left( 1-3i \right)\left( 1+i \right)}{2}=\left( 2-i \right)
⇒1+i−3i−3(i)22=(2−i)\Rightarrow \dfrac{1+i-3i-3{{\left( i \right)}^{2}}}{2}=\left( 2-i \right)
⇒1−2i+32=(2−i)\Rightarrow \dfrac{1-2i+3}{2}=\left( 2-i \right)
⇒4−2i2=(2−i)\Rightarrow \dfrac{4-2i}{2}=\left( 2-i \right)
⇒(2−i)=(2−i)\Rightarrow \left( 2-i \right)=\left( 2-i \right)
LHS = RHS
Here, we get LHS = RHS, hence our value of z is correct.