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Mathematics Question on complex numbers

Let O be the origin and A be the point z1=1+2iz_1 = 1 + 2i. If B is the point z2,Re(z2)<0z_2, Re(z_2) < 0, such that OAB is a right angled isosceles triangle with OB as hypotenuse, then which of the following is NOT true?

A

arg z2=π–tan−13arg\ z_2=\pi–tan^{−1}3

B

(arg)(z1−2z2)=−tan−1⁡43(arg)(z_1−2z_2)=−tan^{−1⁡}\frac 43

C

∣z2∣=10|z_2|=\sqrt {10}

D

∣2z1−z2∣=5|2z_1−z_2|=5

Answer

∣2z1−z2∣=5|2z_1−z_2|=5

Explanation

Solution

Let O be the origin and A be the point z1 = 1 + 2i. If B is the point z2,

z2−0(1+2i)−0=∣OB∣∣OA∣eiπ4\frac {z2−0}{(1+2i)−0}=\frac {|OB|}{|OA|}e^{\frac {i\pi}{4}}

⇒z21+2i=2eiπ4⇒ \frac {z2}{1+2i}=\sqrt 2 e^{i\pi}{4}

z2=(1+2i)(1+i)z_2=(1+2i)(1+i)
z2=−1+3iz_2=−1+3i
arg z2=π–tan−13arg\ z_2=π–tan^{−1}3
∣z2∣=10|z_2|=\sqrt {10}
z1–2z2=(1+2i)+2–6iz_1–2z_2=(1+2i)+2–6i
z1–2z2=3–4iz_1–2z_2=3–4i
arg (z1−2z2)=−tan−1⁡43arg\ (z_1−2z_2)=−tan^{−1⁡}\frac 43
∣2z1−z2∣=∣2+4i+1−3i∣|2z_1−z_2|=|2+4i+1−3i|
∣2z1−z2∣=∣3+i∣|2z_1−z_2|=|3+i|
=10=\sqrt {10}

So, the correct option is (D): ∣2z1−z2∣=5|2z_1−z_2|=5