Question Video: Converting Complex Numbers from Algebraic to Exponential Form | Nagwa Question Video: Converting Complex Numbers from Algebraic to Exponential Form | Nagwa

# Question Video: Converting Complex Numbers from Algebraic to Exponential Form Mathematics • Third Year of Secondary School

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Given that π§ = (πβ(2))/(1 β π), write π§ in exponential form.

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### Video Transcript

Given that π§ equals π root two over one minus π, write π§ in exponential form.

To answer this question, we have two options. We could divide these two complex numbers in algebraic form. And to do this, we need to multiply both the numerator and denominator by the conjugate of the denominator then distribute and simplify as far as possible. Iβm sure youβll agree thatβs rather a lengthy process. Instead, weβll choose to write these complex numbers in exponential form. So weβll need to calculate their moduli and arguments.

π root two is a purely imaginary number. On an Argand diagram, itβs represented by the point whose Cartesian coordinates are zero, root two. Its modulus is the length of the line segment that joins this point to the origin. So itβs root two. And since the argument is measured in the counterclockwise direction from the positive real axis, we can see that the argument of this complex number is equivalent to 90 degrees. Thatβs π by two radians. And in exponential form, we can say that this is the same as root two π to the π by two π.

The complex number one minus π is a little more tricky. Its real part is positive and its imaginary part is negative. So it lies in the fourth quadrant. Now, its modulus is independent of this fact. We simply use the formula the square root of the sum of the square of the real and imaginary parts. So thatβs the square root of one squared plus negative one squared which once again is the square root of two.

We do need to be a little bit more careful with the argument. Since itβs in the fourth quadrant, we can use the formula thatβs unique to complex numbers that are plotted in the first and fourth quadrants. Thatβs arctan of π over π, arctan of the imaginary part divided by the real part. So in this case, thatβs the arctan of negative one over one which is negative π by four. We expected a negative value for the argument as this time weβre measuring in a clockwise direction. And so, we can rewrite our fraction as root two π to the π by two π over root two π to the negative π by four π. And now, we can easily divide.

To divide complex numbers in exponential form, we divide their moduli and subtract their arguments. Root two divided by root two is one and π by two minus negative π by four is three π by four. In exponential form then, π§ is equal to π to the three π by four π.

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