### Video Transcript

Find the complex number π§ that satisfies the equations π§ plus π§ star equals
negative five, and π§ star minus π§ equals three π.

We begin by recalling that a complex number π§ can be written as π§ equals π plus
ππ, where π and π are real, π is the real part of π§, π is the imaginary part,
and π is the square root of negative one. Then, the complex conjugate of π§, that is π§ star, is equal to π minus ππ. So to find π§ star, we simply change the sign of the imaginary part of π§.

Now, we could answer this question in a couple of different ways. One way is to treat these as a pair of simultaneous equations. Calling the first equation equation one and the second equation equation two, we note
that weβve rearranged the left-hand side so that like is above like in the two
equations. So, the π§-term in equation one is above the π§-term in equation two, and π§ star is
above π§ star.

So now summing equations one and two, we have π§ plus negative π§ plus two π§ star is
equal to negative five plus three π. Since π§ plus negative π§ is equal to zero, we have two π§ star is equal to negative
five plus three π. So now dividing through by two, we have π§ star equal to negative five over two plus
three over two π.

So weβve found the complex conjugate π§ star, but weβre asked to find π§ itself. And to do this, we simply need to change the sign of the imaginary part of π§
star. The imaginary part of π§ star is positive three over two, so the imaginary part of π§
is negative three over two. Hence, π§ is equal to negative five over two minus three over two π.

We can solve this in a different way, using some of the properties of a complex
number and its conjugate. One such property is that the sum of π§ with its complex conjugate π§ star equals two
times the real part of π§, and thatβs two π. We also have that the difference between a complex number and its conjugate is two π
times the imaginary part of π§. Now at this point, we can note that the left-hand side of our second equation is
actually the negative of this; thatβs π§ star minus π§. Hence, swapping the order of the difference in our second property, we have π§ star
minus π§ is equal to negative two π times the imaginary part of π§; thatβs negative
two ππ.

So now, we can use these properties to find the real and imaginary parts of the
complex number that satisfies both equations. If π§ is of the form π plus ππ, then from the first property, the sum of π§ with
its conjugate, we have two times the real part of π§, thatβs two π, equals negative
five. And dividing both sides by two gives π equal to negative five over two.

Next applying the difference property to our second equation to find the imaginary
part π, we have negative two ππ is equal to three π. And dividing both sides by negative two π and taking the negative sign up top on the
right, we have π equals negative three over two. And so as expected, we get the same answer for π§, and thatβs π§ equals negative five
over two minus three over two π.