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Question: How do you write the complex number \(6 - 8i\) in polar form?...

How do you write the complex number 6−8i6 - 8i in polar form?

Explanation

Solution

According to the question, we have to write the complex number 6−8i6 - 8i in polar form that is z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i\sin \theta )
So, first of all we have to find rr and θ\theta with the help of the formula mentioned below.

Formula used:
⇒r=x2+y2....................(A)\Rightarrow r = \sqrt {{x^2} + {y^2}} ....................(A)
⇒θ=tan⁡−1(yx)............................(B)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)............................(B)
where xx and yy are the real and imaginary parts respectively of the given complex number 6−8i6 - 8i that is in the form of z=x+iyz = x + iy.

Complete step-by-step answer:
Step 1: First of all we have to let that z=6−8iz = 6 - 8i
Now, we have to compare this equation z=6−8iz = 6 - 8ito the standard form of the complex number that isz=x+iyz = x + iy.
⇒6−8i=x+iy ⇒x=6 ⇒y=−8  \Rightarrow 6 - 8i = x + iy \\\ \Rightarrow x = 6 \\\ \Rightarrow y = - 8 \\\
Step 2: Now, we have to find the value of rr with the help of the formula (A) which is mentioned in the solution hint.
⇒r=(6)2+(−8)2 ⇒r=36+64 ⇒r=100 ⇒r=10  \Rightarrow r = \sqrt {{{\left( 6 \right)}^2} + {{\left( { - 8} \right)}^2}} \\\ \Rightarrow r = \sqrt {36 + 64} \\\ \Rightarrow r = \sqrt {100} \\\ \Rightarrow r = 10 \\\
Step 3: Now we have to find the value of θ\theta with the help of the formula (B) which is mentioned in the solution hint.
⇒θ=tan⁡−1(−86)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{ - 8}}{6}} \right)
Now, 6−8i6 - 8iis in the 4th quadrant so we must ensure that θ\theta is in the 4th quadrant.
⇒θ=−tan⁡−1(1.332)\Rightarrow \theta = - {\tan ^{ - 1}}\left( {1.332} \right)
Now, we know that tan⁡(0.927)=1.332\tan \left( {0.927} \right) = 1.332
⇒θ=−tan⁡−1[(tan⁡0.927)] ⇒θ=−0.927  \Rightarrow \theta = - {\tan ^{ - 1}}\left[ {\left( {\tan 0.927} \right)} \right] \\\ \Rightarrow \theta = - 0.927 \\\
Step 3: Now, we have to find the value of cos⁡θ\cos \theta and sin⁡θ\sin \theta as mentioned below.
⇒cos⁡θ=cos⁡(−0.927) ⇒cos⁡θ=0.6  \Rightarrow \cos \theta = \cos \left( { - 0.927} \right) \\\ \Rightarrow \cos \theta = 0.6 \\\
And,
⇒sin⁡θ=sin⁡(−0.927) ⇒sin⁡θ=−0.8  \Rightarrow \sin \theta = \sin \left( { - 0.927} \right) \\\ \Rightarrow \sin \theta = - 0.8 \\\
Step 4: Now, the polar form of the given complex number 6−8i6 - 8i is in the form of z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i\sin \theta ) that is z=10(0.6−i0.8)z = 10(0.6 - i0.8)

Final solution: Hence, the polar form of the given complex number 6−8i6 - 8i is 10(0.6−i0.8)10(0.6 - i0.8)

Note:
It is necessary to understand about the complex number and the formulas which are mentioned in the solution hint.
It is necessary to check that the given complex number 6−8i6 - 8i lies in which quadrant.