Question Video: Solving Complex Number Equations and Converting Them from Algebraic to Polar Form | Nagwa Question Video: Solving Complex Number Equations and Converting Them from Algebraic to Polar Form | Nagwa

Question Video: Solving Complex Number Equations and Converting Them from Algebraic to Polar Form Mathematics • Third Year of Secondary School

Given that 𝑧 − 2 = (𝑧 + 2) 𝑖, determine the trigonometric form of the complex number 𝑧.

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Video Transcript

Given that 𝑧 minus two is equal to 𝑧 plus two multiplied by 𝑖, determine the trigonometric form of the complex number 𝑧.

The rectangular or algebraic form of a complex number is 𝑧 is equal to 𝑎 plus 𝑏𝑖. Notice how our equation looks slightly different to this. We’re going to begin by making 𝑧 the subject of our equation. To start, we’ll expand these brackets. 𝑧 multiplied by 𝑖 is 𝑧𝑖. And two multiplied by 𝑖 is two 𝑖. Next, we’ll add two to both sides of this equation. And we get that 𝑧 is equal to 𝑧𝑖 plus two plus two 𝑖. And then we’ll subtract 𝑧𝑖 from both sides.

At this stage, to make 𝑧 the subject, we’re going to need to factorise the left-hand side of the equation. 𝑧 and negative 𝑧𝑖 share a factor of 𝑧. So that goes on the outside of our bracket. 𝑧 divided by 𝑧 is one. And negative 𝑧𝑖 divided by 𝑧 is negative 𝑖. So this factorises to 𝑧 multiplied by one minus 𝑖. And our final stage in making 𝑧 the subject is to divide through by one minus 𝑖.

So we see that 𝑧 is equal to two plus two 𝑖 all over one minus 𝑖. This still doesn’t look quite like the algebraic form we saw earlier. So before we can write it in trigonometric form, we’re going to work out two plus two 𝑖 divided by one minus 𝑖. And to do that, we find the complex conjugate of the denominator.

For the general complex number given as 𝑎 plus 𝑏𝑖, its conjugate is 𝑎 minus 𝑏𝑖. So the conjugate of one minus 𝑖 is one plus 𝑖. And I’m going to multiply both the numerator and the denominator of this fraction by one plus 𝑖. We’ll begin by multiplying the brackets on the top of this expression. Two multiplied by one is two. Two multiplied by 𝑖 is two 𝑖. And two 𝑖 multiplied by one is two 𝑖 again. And then we have two 𝑖 multiplied by 𝑖 which is two 𝑖 squared.

Remember, 𝑖 squared is the square of the square root of negative one. So it’s negative one. So two 𝑖 squared becomes negative two. And we see that two plus two 𝑖 multiplied by one plus 𝑖 is four 𝑖. Let’s repeat this process for the denominator. One multiplied by one is one. We then have one multiplied by 𝑖 which is 𝑖 and negative 𝑖 multiplied by one which is negative 𝑖. Negative 𝑖 multiplied by 𝑖 is negative 𝑖 squared. And remember, 𝑖 squared is negative one. So negative 𝑖 squared becomes one. And our expression simplifies to one plus one which is two.

Our complex number can be written as four 𝑖 over two which is simply two 𝑖. Now that we have this complex number in rectangular form, we recall that to write it in trigonometric form. It’s 𝑟 cos 𝜃 plus 𝑖 sin 𝜃 where 𝑟 is the modulus of the complex number and 𝜃 is its argument.

We generally prefer 𝜃 to be in radians. So we need to find a way to represent the real and imaginary components of our number in terms of 𝑟 and 𝜃. In fact, we can use these formulae to help us. The modulus is the square root of 𝑎 squared plus 𝑏 squared. And to find the value of 𝜃, we use arctan of 𝑏 over 𝑎.

So let’s substitute what we know about our a complex number into these formulae. In the complex number, 𝑧 is equal to two 𝑖. 𝑎 is equal to zero. And 𝑏 is equal to two. So the modulus is the square root of zero squared plus two squared which is simply two. And 𝜃 is arctan of two over zero. Now, two divided by zero is undefined. And 𝜃 could be 𝜋 over two or negative 𝜋 over two. We need to make a decision as to which one we choose. Here, we consider the Argand diagram. On this Argand diagram, our complex number 𝑧 is equal to two 𝑖 is here.

Remember, 𝜃 is the angle made with the real axis. And it’s measured in a counterclockwise direction. So we can see that our value of 𝜃 must be positive 𝜋 over two. A 𝜃 value of negative 𝜋 over two would actually give us negative two 𝑖 or similar. And we therefore see that the trigonometric form of our complex number is two multiplied by cos of 𝜋 over two plus 𝑖 sin of 𝜋 over two.

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