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Sagot :
Our problem involves operation of complex numbers. When dealing with complex number specially imaginary numbers (the one with an "i" ). We must remember that i² = -1.
So let us start.
[tex]\frac{(2-3i)(5i)}{(2+3i)}[/tex]Step 1. Rationalize the denominator by multiplying both the numerator and denominator by (2 - 3i ), in order for us to get rid of the "i" in our denominator.
[tex]\frac{(2-3i)(5i)}{(2+3i)}\cdot\frac{(2-3i)}{(2-3i)}=\frac{(2-3i)(5i)(2-3i)}{(2+3i)(2-3i)}[/tex]*Notice that our denominator is just the SUM and DIFFERENCE of a SQUARE, therefore it will now become,
[tex]\begin{gathered} \frac{(2-3i)(5i)(2-3i)}{(2+3i)(2-3i)}=\frac{(2-3i)(5i)(2-3i)}{2^2-3^2(i^2)}=\frac{(2-3i)(5i)(2-3i)}{4-9(-1)_{}_{}_{}} \\ =\frac{(2-3i)(5i)(2-3i)}{13} \end{gathered}[/tex]STEP 2. Now that we've settled our denominator let us now proceed to the numerator.
[tex]\frac{(2-3i)(5i)(2-3i)}{13}=\frac{(2-3i)^2(5i)}{13}=\frac{(4-12i+9i^2)^{}(5i)}{13}[/tex][tex]=\frac{(4-12i+9(-1)^{})^{}(5i)}{13}=\frac{(-5-12i)^{}(5i)}{13}[/tex]Now let us distribute 5i to the (-5 - 12i),
[tex]\frac{(-5-12i)^{}(5i)}{13}=\frac{(-25i-60i^2)^{}}{13}=\frac{(-25i-60(-1))^{}}{13}[/tex][tex]\frac{(-25i-60(-1))^{}}{13}=\frac{60-25i}{13}[/tex]Therefore the answer for our operation is:
[tex]\frac{60-25i}{13}[/tex]
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