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Question: Check whether it is true or not, if \({z^2} = - 3 + 4i\) then \(z = \pm \left( {1 + 2i} \right)\)?...

Check whether it is true or not, if z2=−3+4i{z^2} = - 3 + 4i then z=±(1+2i)z = \pm \left( {1 + 2i} \right)?

Explanation

Solution

Hint: Here, we will proceed by converting the RHS of the given equation i.e., z2=−3+4i{z^2} = - 3 + 4i in such a ways that the RHS is in the form of a2+b2+2(a)(b){a^2} + {b^2} + 2\left( a \right)\left( b \right) so that we can use the formula a2+b2+2(a)(b)=(a+b)2{a^2} + {b^2} + 2\left( a \right)\left( b \right) = {\left( {a + b} \right)^2}.

Complete step-by-step answer:
Given, z2=−3+4i{z^2} = - 3 + 4i
We can rewrite the above equation by replacing -3 with (-4+1) as given under
z2=−4+1+4i ⇒z2=(−1)4+1+4i  {z^2} = - 4 + 1 + 4i \\\ \Rightarrow {z^2} = \left( { - 1} \right)4 + 1 + 4i \\\
As we know that i2=−1{i^2} = - 1, the above equation becomes
⇒z2=(i2)4+1+4i ⇒z2=4i2+1+4i  \Rightarrow {z^2} = \left( {{i^2}} \right)4 + 1 + 4i \\\ \Rightarrow {z^2} = 4{i^2} + 1 + 4i \\\
We can rewrite the above equation by replacing 4i24{i^2} by (2i)2{\left( {2i} \right)^2}, we have
⇒z2=(2i)2+1+4i\Rightarrow {z^2} = {\left( {2i} \right)^2} + 1 + 4i
We can rewrite the above equation by replacing 1 by (1)2{\left( 1 \right)^2} and 4i4i by 2(2i)(1)2\left( {2i} \right)\left( 1 \right) , we have
⇒z2=(2i)2+(1)2+2(2i)(1)\Rightarrow {z^2} = {\left( {2i} \right)^2} + {\left( 1 \right)^2} + 2\left( {2i} \right)\left( 1 \right)
Using the formula a2+b2+2(a)(b)=(a+b)2{a^2} + {b^2} + 2\left( a \right)\left( b \right) = {\left( {a + b} \right)^2}, we get
⇒z2=(2i+1)2 ⇒z2=(1+2i)2  \Rightarrow {z^2} = {\left( {2i + 1} \right)^2} \\\ \Rightarrow {z^2} = {\left( {1 + 2i} \right)^2} \\\
By taking square root on both sides, we get
⇒z=±(1+2i)2 ⇒z=±(1+2i)  \Rightarrow z = \pm \sqrt {{{\left( {1 + 2i} \right)}^2}} \\\ \Rightarrow z = \pm \left( {1 + 2i} \right) \\\
Therefore, if z2=−3+4i{z^2} = - 3 + 4i then z=±(1+2i)z = \pm \left( {1 + 2i} \right) is true.

Note: In this particular problem, the most important step is z=±(1+2i)2z = \pm \sqrt {{{\left( {1 + 2i} \right)}^2}} where we can clearly see that after taking the square root on both the sides of the equation z2=(1+2i)2{z^2} = {\left( {1 + 2i} \right)^2}, we have considered both positive as well as negative signs. So, the result is z=±(1+2i)z = \pm \left( {1 + 2i} \right) which means that the complex number z has two values which are z=1+2iz = 1 + 2i and z=−1−2iz = - 1 - 2i.