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Question: If \({{z}_{1}}\) and \({{z}_{2}}\) are two complex numbers such that \(\left| {{z}_{1}}+{{z}_{2}} \r...

If z1{{z}_{1}} and z2{{z}_{2}} are two complex numbers such that ∣z1+z2∣=∣z1−z2∣\left| {{z}_{1}}+{{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right| then Arg z1−Arg z2=Arg\,{{z}_{1}}-Arg\,{{z}_{2}}=
A. 00
B. π2\dfrac{\pi }{2}
C. π3\dfrac{\pi }{3}
D. π4\dfrac{\pi }{4}

Explanation

Solution

Firstly write zz in the complex number form and then substitute it in the given condition, ∣z1+z2∣=∣z1−z2∣\left| {{z}_{1}}+{{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right|. For getting rid of the i, square the equation on both sides or use the distance formula. Now we shall here get a certain condition which will be later used to help in solving Arg z1−Arg z2Arg\,{{z}_{1}}-Arg\,{{z}_{2}} . Now substitute the value of Arg z=tan⁡−1(yx)Arg\,z={{\tan }^{-1}}\left( \dfrac{y}{x} \right) and then find the angle.

Complete step by step solution:
Firstly, we represent zz in complex number form.
z1{{z}_{1}} can be written as x1+iy1{{x}_{1}}+i{{y}_{1}}
z2{{z}_{2}} can be written as x2+iy2{{x}_{2}}+i{{y}_{2}}
Now let us put this in the given statement.
⇒∣z1+z2∣=∣z1−z2∣\Rightarrow \left| {{z}_{1}}+{{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right|
Now substitute the values of z1,z2{{z}_{1}},{{z}_{2}} in the expression.
⇒∣x1+iy1+x2+iy2∣=∣x1+iy1−x2−iy2∣\Rightarrow \left| {{x}_{1}}+i{{y}_{1}}+{{x}_{2}}+i{{y}_{2}} \right|=\left| {{x}_{1}}+i{{y}_{1}}-{{x}_{2}}-i{{y}_{2}} \right|
Now group the like variables together and take the commons out.
⇒∣x1+x2+i(y1+y2)∣=∣x1−x2+i(y1−y2)∣\Rightarrow \left| {{x}_{1}}+{{x}_{2}}+i\left( {{y}_{1}}+{{y}_{2}} \right) \right|=\left| {{x}_{1}}-{{x}_{2}}+i\left( {{y}_{1}}-{{y}_{2}} \right) \right|
Now represent the expression in the distance formula.
⇒(x1+x2)2+(y1+y2)2=(x1−x2)2+(y1−y2)2\Rightarrow \sqrt{{{\left( {{x}_{1}}+{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}+{{y}_{2}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}
Now upon squaring on both sides, we get,
⇒x12+x22+2x1x2+y12+y22+2y1y2=x12+x22−2x1x2+y12+y22−2y1y2\Rightarrow {{x}_{1}}^{2}+{{x}_{2}}^{2}+2{{x}_{1}}{{x}_{2}}+{{y}_{1}}^{2}+{{y}_{2}}^{2}+2{{y}_{1}}{{y}_{2}}={{x}_{1}}^{2}+{{x}_{2}}^{2}-2{{x}_{1}}{{x}_{2}}+{{y}_{1}}^{2}+{{y}_{2}}^{2}-2{{y}_{1}}{{y}_{2}}
Upon canceling the common terms on both sides of the expression we get,
⇒2x1x2+2y1y2=−2x1x2−2y1y2\Rightarrow 2{{x}_{1}}{{x}_{2}}+2{{y}_{1}}{{y}_{2}}=-2{{x}_{1}}{{x}_{2}}-2{{y}_{1}}{{y}_{2}}
On further evaluating we get,
⇒4x1x2+4y1y2=0\Rightarrow 4{{x}_{1}}{{x}_{2}}+4{{y}_{1}}{{y}_{2}}=0
We can rewrite this as,
⇒x1x2+y1y2=0\Rightarrow {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}=0
Now shall solve, arg⁡z1−arg⁡z2\arg {{z}_{1}}-\arg {{z}_{2}}
We can write arg⁡z1−arg⁡z2\arg {{z}_{1}}-\arg {{z}_{2}} as arg⁡(z1z2)\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)
Now let us again substitute the values ofz1,z2{{z}_{1}},{{z}_{2}} in complex number form.
⇒arg⁡(x1+iy1x2+iy2)\Rightarrow \arg \left( \dfrac{{{x}_{1}}+i{{y}_{1}}}{{{x}_{2}}+i{{y}_{2}}} \right)
Now let us rationalize the denominator.
Upon rationalizing we get,
⇒arg⁡(x1+iy1x2+iy2×x2−iy2x2−iy2)\Rightarrow \arg \left( \dfrac{{{x}_{1}}+i{{y}_{1}}}{{{x}_{2}}+i{{y}_{2}}}\times \dfrac{{{x}_{2}}-i{{y}_{2}}}{{{x}_{2}}-i{{y}_{2}}} \right)
Now evaluate this expression.
⇒arg⁡((x1+iy1)(x2−iy2)x22+y22)\Rightarrow \arg \left( \dfrac{\left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}-i{{y}_{2}} \right)}{{{x}_{2}}^{2}+{{y}_{2}}^{2}} \right)
Now simplify the numerator part.
⇒arg⁡(x1x2−iy2x1+iy1x2+y1y2x22+y22)\Rightarrow \arg \left( \dfrac{{{x}_{1}}{{x}_{2}}-i{{y}_{2}}{{x}_{1}}+i{{y}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}{{{x}_{2}}^{2}+{{y}_{2}}^{2}} \right)
Now rearrange the terms.
⇒arg⁡(x1x2+y1y2+i(y1x2−y2x1)x22+y22)\Rightarrow \arg \left( \dfrac{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+i\left( {{y}_{1}}{{x}_{2}}-{{y}_{2}}{{x}_{1}} \right)}{{{x}_{2}}^{2}+{{y}_{2}}^{2}} \right)
Now upon writing this in the form, Arg z=tan⁡−1(yx)Arg\,z={{\tan }^{-1}}\left( \dfrac{y}{x} \right)
We get,
⇒tan⁡−1(x1y2−x2y1x1x2+y1y2)\Rightarrow {{\tan }^{-1}}\left( \dfrac{{{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}}}{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}} \right)
But we have proved that,
x1x2+y1y2=0{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}=0
Now on substituting that we get,
⇒tan⁡−1(x1y2−x2y10)\Rightarrow {{\tan }^{-1}}\left( \dfrac{{{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}}}{0} \right)
We know that anything divided by zero is infinity.
⇒tan⁡−1(∞)\Rightarrow {{\tan }^{-1}}\left( \infty \right)
This is equal to π2\dfrac{\pi }{2}
Hence, Argz1−Argz2Arg{{z}_{1}}-Arg{{z}_{2}} when ∣z1+z2∣=∣z1−z2∣\left| {{z}_{1}}+{{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right| is equal to π2\dfrac{\pi }{2}

So, the correct answer is “Option B”.

Note: The inverse of tan⁡\tan is nothing but arctan⁡\arctan. The inverse functions in trigonometry are also known as arc functions or anti trigonometric functions. They are majorly known as arc functions because they are most used to find the length of the arc needed to get the given or specified value. We can convert a function into an inverse function and vice versa.