### Video Transcript

Without first converting π§ to
trigonometric form, determine the square roots of π§, where π§ equals negative eight
add two π over one add four π.

Before we begin trying to solve
this problem, letβs write our complex number π§ in the form π plus ππ, where π
is the real part and π is the imaginary part. To do this, weβll divide the
complex number negative eight add two π by one add four π. We follow the usual procedure by
using the conjugate of the denominator, that is, one minus four π. So we multiply π§ by one minus four
π over one minus four π, which is essentially just multiplying by one. So weβre not changing π§ in any
way.

Next, we multiply the numerators
together. That gives us negative eight minus
eight π squared plus two π plus 32π. And then, we multiply the
denominators together. That gives us one minus 16π
squared plus four π minus four π. But now remember that by
definition, π squared is equal to negative one. So where we see π squared, we can
replace this with negative one. And now, we can just simplify this
a little bit. Starting with the numerator, eight
multiplied by negative one is negative eight. But because weβve already got a
minus at the front, this becomes a plus eight. And then, negative eight add eight
is just going to give us zero.

We can also gather together like
terms. Two π add 32π will give us
34π. Then for the denominator, 16
multiplied by negative one will give us negative 16. But because thereβs already a minus
at the front, this is plus 16. And four π subtract four π will
just give us zero. So we end up with 34π over 17. But this just gives us two π.

So now weβve rewritten π§ as two
π, we can now proceed to find its square roots. Iβm just gonna clear some space so
that we can do this. Letβs say the square root of two π
equals the complex number π plus ππ. Weβre going to find the values for
π and π. Starting off by squaring both sides
gives us two π equals π squared add πππ add πππ add π squared π
squared. We can then simplify this to π
squared add two πππ add π squared π squared. But remember by definition, π
squared is equal to negative one. So we can write this as minus π
squared.

In order to solve this for π and
π, we can group these terms together, like so. Then, we can compare the terms on
the left of the expression with the terms on the right of the expression. On the left of the expression, we
have no real terms. So we could say that the real part
is zero. And on the right of the expression,
the real part is π squared minus π squared. So this gives us that π squared
minus π squared is equal to zero. And now comparing the imaginary
part on the left with the imaginary part on the right, the imaginary part on the
left is two. And the imaginary part on the right
is two ππ. So that gives us that two ππ must
be equal to two. But we can actually cancel
this. Two ππ equals one.

So now, we have two expressions
involving π and π. Weβre going to use these two
expressions now to solve for π and π. From equation two, we have that π
equals one over π. So letβs substitute this into
equation one. This gives us π squared minus one
over π squared equals zero. Letβs now evaluate one over π
squared. This is going to give us π squared
minus one over π squared equals zero, because we just square the numerator and we
square the denominator.

To proceed from here, letβs
multiply through by π squared. That gives us π to the fourth
power minus one equals zero. And to solve for π, we can factor
this expression as π squared add one multiplied by π squared minus one equals
zero. This gives us the solutions of π
squared equals negative one and π squared equals one. π squared equals one gives us that
π is positive or negative one and π squared equals negative one gives us that π
equals positive or negative π.

So now we found values for π,
weβve got to find the corresponding values for π. Letβs take π to be positive or
negative one. When π equals one, π equals one
over one. This is just using the expression
that we found earlier for π. And one over one is just one. When π equals negative one, π
equals one over negative one, which is just negative one. So then, the complex number which
we said is our solution to the square root of two π, π plus ππ. When π is one and π is one, this
is one plus π. And when π is negative one and π
is negative one, this is negative one minus π.

As weβre looking for a square root,
weβre going to have two solutions. So these are our two solutions: one
plus π and negative one minus π. But you might be wondering about
the two other values we found for π, positive and negative π. Well actually, if we take π to be
π, then π would be one over π. And if we take π to be negative
π, then π would be one over negative π or just negative one over π.

When π equals π and π equals one
over π, π plus ππ is π plus one over π multiplied by π. But thatβs just π plus one. And when π is negative π and π
is negative one over π, π plus ππ is negative π plus negative one over π
multiplied by π. But that just gives us negative π
minus one. And actually, we see that these are
exactly the same solutions that we just found. So the square roots of π§ are one
plus π and negative one minus π.