Question

The conjugate of complex number 2-3i/4-i​

Answer 1

$\dfrac{11}{17}+\dfrac{10}{17}i$

The conjugate of the complex number $\dfrac{2-3i}{4-i}$ is $\dfrac{11}{17}+\dfrac{10}{17}i$.

Given data:

The complex number $\dfrac{2-3i}{4-i}$

To find:

The conjugate complex number

Concept:

• If $a+ib$ be a complex number, then its conjugate is $a-ib$.

• If $a-ib$ be a complex number, then its conjugate is $a+ib$.

Step-by-step explanation:

Here, the given complex number is

$\dfrac{2-3i}{4-i}$

= $\dfrac{(2-3i)(4+i)}{(4-i)(4+i)}$

• We have multiplied both the numerator and the denominator by the conjugate of the complex number in the denominator.

= $\dfrac{8+2i-12i-3i^{2}}{16-i^{2}}$

= $\dfrac{8-10i+3}{16+1}$

• since $i=\sqrt{-1}\Rightarrow i^{2}=-1$

= $\dfrac{11-10i}{17}$

= $\dfrac{11}{17}-\dfrac{10}{17}i$

So, the required conjugate is

$\dfrac{11}{17}+\dfrac{10}{17}i$

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